Bruce H. Edwards
Complete Solutions
Guide
Volume
. X^iKT''-^'^',*
Calculus
Seventh Edition
^jl i- •' ■■^"WJ :.--■?':'
Larson • Hostetler • Edwards
For use with
Calculus with Analytic Geometry, Seventh Edition
Calculus of a Single Variable, Seventh Edition
Digitized by the Internet Archive
in 2011 with funding from
Lakewood Park Christian School
http://www.archive.org/details/calculuscomplete01edwa
PART I
CHAPTER P
Preparation for Calculus
Section P.l Graphs and Models 2
Section P.2 Linear Models and Rates of Change 7
Section R3 Functions and Their Graphs 14
Section P.4 Fitting Models to Data 18
Review Exercises 19
Problem Solving 23
CHAPTER P
Preparation for Calculus
Section P.l Graphs and Models
Solutions to Odd- Numbered Exercises
1. >> = —jx + 2
x-intercept: (4, 0)
y-intercept: (0, 2)
Matches graph (b)
3. y = 4 - x^
x-intercepts: (2,0), (-2,0)
y-intercept: (0, 4)
Matches graph (a)
5. >> = 5X + 1
7. >- = 4 - x2
X
-4
-2
0
2
4
y
-5
-2
1
4
7
X
-3
-2
0
2
3
y
-5
0
4
0
-5
(3,-5)
9. .V = U + 21
X
-5
-4
-3
-2
-1
0
1
y
3
2
1
0
1
2
3
11. y = 7x - 4
X
0
1
4
9 16
>"
-4
-3
-2
-1 0
Section P. I Graphs and Models 3
13.
Xmin = -3
Xmax = 5
Xscl = 1
Ymin = -3
Ymax = 5
Yscl = 1
Note that y = A when x = 0.
15.
(-4.f«. 3,
i:. 1.73)
(a) (2,.v) = (2, 1.73) (y = V5 - 2 = 73 = 1.73)
(b) (x, 3) = (-4, 3) (3 = 75 - (-4))
\1. y = x^ + X- 2
y-intercept: y = 0"^ + 0 — 2
y=-2;(0, -2)
0 = x~ + X - 2
jc-intercepts:
0 = (x + 2){x - 1)
X = -2, 1; (-2,0), (1,0)
21. V =
3(2 - V^)
y-intercept: None, x cannot equal 0.
JT-intercepts:
0 = 2 - Vx
.1 = 4: (4, 0)
19. y = xV25 - .t'
y-intercept: y = 0^725 - 0^
y = 0; (0, 0)
jc-intercepts:
; 0 = .r=v/25^^
0 = .r=7(5 - .x)(5 + x)
x = 0, ±5: (0,0): (±5,0)
23. .r^' - x^ + Ay = 0
y-intercept:
O^Cv) - 02 -(- 4y = 0
y = 0: (0. 0)
-T-intercept:
x-{Q) - X- + 4(0) = 0
x = Q; (0, 0)
25. Symmetric with respect to the y-axis since
y = (— x)- — 2 = X- — 2.
27. Symmetric with respect to the x-axis since
(-y)- = y^ = x^ - 4x.
29. Symmetric with respect to the origin since
(-x)(-y) = xy = 4.
31. y = 4 - 7x-l- 3
No symmetry with respect to either axis or the origin.
33. Symmetric with respect to the origin since
—X
' i-xf + 1
X
^ x^ + r
37.
y=-3x + 2
Intercepts:
(i0),(0,2)
Symmetry: none
35. y = |.T^ + x| is symmetric with respect to the y-axis
since y = \(-xy + (-.r)| = \-(x^ + x)\ = \x^ + x\.
4 Chapter P Preparation for Calculus
39. y = |.v - 4
Intercepts:
(8,0), (0,-4)
Symmetry: none
45. )> = -x^ + 2
Intercepts:
(-3^,0), (0,2)
Symmetry: none
\ 1 \-^x
41. y = 1 - x^
Intercepts:
(1,0), (-1,0), (0,1)
Symmetry: y-axis
47. y = xjx + 2
Intercepts:
(0,0), (-2,0)
Symmetry: none
Domain: x > ~2
43. y = (x + 3)2
Intercepts:
(-3,0), (0,9)
Symmetry: none
-10 -8 -6 (-3,0)
49. ;c = /
Intercepts: (0,0)
Symmetry: origin
Intercepts: none
Symmetry: origin
53. y = 6 - |.y|
Intercepts:
(0,6), (-6,0), (6,0)
Symmetry: >'-axis
55. y^
- x=9
y^ = x + 9
y = ±Jx + 9
Intercepts:
(0,3),(0, -3), (-9,0)
Symmetry: jr-axis
57. X + 3y2 = 6
3y2 = 6 - ;t
Intercepts:
(6, 0), (O, V2), (O, - V2)
Symmetry: A-axis
Section p.] Graphs and Models 5
59. y = (x + 2){x - 4){x - 6) (other answers possible)
61. Some possible equations:
y = x
y = x^
y = 3x^ — X
„ = 3/;
63. x+y = 2=>y = 2-x
2x — y = I =^ y = 2x — 1
2 -;c = 2x- 1
3 = 3;c
1 =x
The corresponding >>- value \sy = 1.
Point of intersection: (1,1)
65. X + y = 1 =^ y = 1 — X
Zx - 11
\A-lx='ix- 11
-5x= -25
;c = 5
The corresponding >>- value is >> = 2.
Point of intersection: (5, 2)
61. x'^+y = 6^y = 6- x'^
x + y = A=!>y = A — X
6- x^ = A- X
0 = x^ - x~2
Q = {x- 2)(x + 1)
;c = 2, - 1
The corresponding y- values are y = 2 (for x = 2)
and y = 5 (for x = — 1 ).
Points of intersection: (2, 2), (—1,5)
69. ;c2 + y2 = 5 => y2 = 5 - ;c2
jr — y= \ => y = X — 1
5- x- = {x- 1)2
5-x^ = x--2x+\
Q = 2x~ -2x- A = 2(x+ \){x - 2)
X = - 1 or .r = 2
The corresponding y- values are y - - 2 and y = 1 .
Points of intersection: (- 1, -2), (2. 1)
71. y = x3
y = X
x? = X
:^ - x = Q
x{x + \)(x - 1) = 0
X = Q,x = — 1, orx = 1
The corresponding y- values are v = 0, y = - 1, and
y=l.
Points of intersection: (0, 0), (- 1, - 1), (1, 1)
73.
>• =
x> -Ztt + x- \
y = -jr + 3.t - 1
^ -2x- + x- \ = -X- + 3.r - 1
.r3 - .^2 - It = 0
x(x - 2)(x + 1) = 0
.r= -1,0.2
(-1, -5),(0. -1).(2. 1)
l,^^:dt
7^
^,2.U
(O.-in
\
-
' 1 '"^^
3
6 Chapter P Preparation for Calculus
75. 5.5v^ + 10,000 = 3.29x
(5.5^)^ = (3.29x - 10,000)2
30.25;c = 10.8241JC2 - 65,800x + 100,000,000
0 = 10.8241^2 - 65,830.25;c + 100,000,000 Use the Quadratic Formula.
X = 3133 units
The other root, x = 2949, does not satisfy the equation R = C.
This problem can also be solved by using a graphing utility and finding the intersection of the graphs of C and R.
77. (a) Using a graphing utility, you obtain
y = -0.0153f- + 4.9971f + 34.9405
(c) For the year 2004. J = 34 and
y = 187.2 CPl.
(b)
79. 400
If the diameter is doubled, the resistance is changed by approximately a factor of (1/4). For instance, ^(20) ~ 26.555 and
. ^(40) = 6.36125.
81. False; j:-axis symmetry means that if (1, -2) is on the graph, then (1, 2) is also on the graph.
83. True; the x-intercepts are
-b± JlP- - 4ac
2a
, 0
85. Distance to the origin = K x Distance to (2, 0)
Jx- + y^ = A:V(x - 2)2 + f, K + 1
x^ + y'^ = K-{x^ - Ax + A + y-)
(1 - K'^)x^ + (1 - K^)y^ + 4Kh - 4K^ = 0
Note: This is the equation of a circle!
Section P.2 Linear Models and Rates of Change 7
Section P.2 Linear Models and Rates of Cliange
\. m=\
3. m = 0
5. m= -\2
7,
9. m
2 - (-4)
5-3
(3, -4)
11. m
5 - 1
2-2
^4
0
undefined
(2,5)
(2,1)
13. m =
2/3 - 1/6
1/2 -(-3/4)
111
1/4
15. Since the slope is 0, the line is horizontal and its equation is >■ = 1. Therefore, three additional points are (0, 1), (1, 1),
and (3, 1).
17. The equation of this line is
y-l = -3{x-\)
y= -3x + 10.
Therefore, three additional points are (0, 10), (2, 4), and (3, 1).
19. Given a line L, you can use any two distinct points to calculate its slope. Since a line is straight, the ratio of the change in
y-values to the change in jr-values will always be the same. See Section P.2 Exercise 93 for a proof.
8 Chapter P Preparation for Calculus
21. (a)
E
= 260- ■
I I I I I I I I l>
123456789
Year (0<-» 1990)
(b) The slopes of the line segments are
255.0 - 252.1
2 - 1
257.7 - 255.0
3-2
260.3 - 257.7
4-3
262.8 - 260.3
5-4
265.2 - 262.8
6-5
267.7 - 265.2
7-6
270.3 - 267.7
8-7
2.9
2.7
= 2.6
= 2.5
2.4
2.5
2.6
The population increased most rapidly from 1991 to 1992.
(m = 2.9)
23. x + 5y--20
Therefore, the slope is /n =
(0,4).
and the y-intercept is
25. ;c = 4
The line is vertical. Therefore, the slope is undefined and
there is no y-intercept.
27. y = |.r + 3
Ay = -ix+ 12
0 = 3;c - 4y + 12
31. y + 2 = 3(;c - 3)
y + 2 = 3x - 9
y = 3.r - 11
y - 3a:+ 11 = 0
33. m
6-0
= 3
2-0
y - 0 = 3U - 0)
y = 3x
35. m
1 - (-3)
37. m
8
-0
8
2
- 5
3
y -
0
= -1-
-5)
y
= -|-
40
3
3y + 8jc - 40 = 0
Section P.2 Linear Models and Rates of Change 9
39. m
1
Undefined.
5-5
Vertical line jr = 5
(5.8)
(5.1)
41. m
1/1 - 3/4 11/4 _ 11
1/2
- 0
1/2
y-
3 _
4
^U - 0)
y =
11 ^3
T^ + i
llx' Ay + ?, = Q
43. .r = 3
jc- 3 = 0
(3,0)
45. ^ + f=l
2 3
3;c + 2j' - 6 = 0
47.^ + ^=1
a a
1 + ^=1
a a
- = 1
a
a = 3=»jr + y = 3
.X + V - 3 = 0
49. y = -3
y + 3 = 0
12 3 4 5
51. y=-lx+\
53. y - 2 = fU - 1)
y = 2.t + 2
2y - 3x - 1 = 0
55. 2t - y - 3 = 0
y = 2x - 3
10 Chapter P Preparation for Calculus
57.
The lines do not appear perpendicular.
The lines appear perpendicular.
The lines are perpendicular because their slopes 1 and — 1 are negative reciprocals of each other.
You must use a square setting in order for perpendicular lines to appear perpendicular.
59. 4a: - ly = 3
y = 2x - 2
/n = 2
(a) y - 1 = 2{x - 2)
y - \ =2x- 4
2x - y - 3 = 0
(b) J - 1 = -\{x - 2)
2>' - 2 = -;c + 2
X + 2y - A = 0
61, 5;c - 3y = 0
y = h
m = l
(a) y-l
= 1(^-1)
24y - 21
= 40;c - 30
24y - 40jc + 9
= 0
(b) y-l
= -f(- - !)
40y - 35
= -24a; + 1
40y + 24;c - 53
= 0
63. (a) x = 2 => X - 2 = 0
0})y = 5=>y-5 = 0
65. The slope is 125. Hence, V = 125(f - 1) + 2540
= 125? + 2415
67. The slope is -2000. Hence, V
-2000(f - 1) + 20,400
-2000/ + 22,400
69.
You can use the graphing utility to determine that the points of intersection are (0, 0) and (2, 4). Analytically,
x^ = 4x - x^
2x2 - 4x = 0
2x{x - 2) = 0
x = 0=^y = 0^>{0,0)
X = 2 => >- = 4 ^ (2, 4).
The slope of the line joining (0, 0) and (2, 4) is m = (4 - 0)/(2 — 0) = 2. Hence, an equation of the line is
>- - 0 = 2(x - 0)
y = 2x.
Section P.2 Linear Models and Rates of Change 11
71. m, =
1 -0
-2-(-l)
-2- 0 _ _2
"^"l-C-D" 3
The points are not collinear.
73. Equations of perpendicular bisectors:
c a — b
c a + b
a + b
b- a
Letting ;<: = 0 in either equation gives the point of inter-
section:
0,
2c j
This point lies on the third perpendicular bisector, ^ = 0.
(-a,0)
75. Equations of altitudes:
a- b,
-ix + a)
x = b
a + b. ,
y = U - a)
Solving simultaneously, the point of intersection is
.2_ ^\
[-'-
77. Find the equation of the line through the points (0, 32) and (100, 212).
180 9
"^ ~ 100 ~ 5
f - 32 = f (C - 0)
f = f C + 32
5F - 9C - 160 = 0
For F = 72°, C = 22.2°.
79. (a) W, = 0.75j: + 12.50
W. = 1.30.t + 9.20
(c) Both jobs pay $17 per hour if 6 units are produced.
For someone who can produce more than 6 units per
hour, the second offer would pay more. For a worker
who produces less than 6 units per hour, the first oifer
pays more.
(b) 50
Using a graphing utility, the point of intersection is
approximately (6. 17). Analytically,
0.75.x- + 12.50 = 1.30.V + 9.20
3.3 = 0.55.V => .V = 6
y = 0.75(6) + 12.50 = 17.
12 Chapter P Preparation for Calculus
81. (a) Two points are (50, 580) and (47, 625). The slope is
625-580
'"= 47-50 =-''■
p - 580 = - \5{x - 50)
p= - \5x + 750 + 580 = - \5x + 1330
or .r = 15(1330 - p)
(b) =0
Ifp = 655, ;t = -^(1330 - 655) = 45 units,
(c) \£p = 595, X = n(1330 - 595) = 49 units.
83. 4r + 3v - \Q = Q^>d =
_ |4(0) + 3(0) - 10| _ 10 _
J A- + r-
i5. x-y-2 = Q:
|l(-2) + (-l)(l)-2| ^^
VI- + r- V2
5./2
87. A point on the line x + y = 1 is (0, 1). The distance from the point (0, 1) to x + y — 5 = 0 is
11(0) + 1(1) -51 11-51 4 ,-
d =
JV- + 1-
v/2 ./2
89. If A = 0, then fiy + C = 0 is the horizontal line y = — C/B. The distance to {x^, yj is
|5y, + C\ JAri + By, + C\
d =
yi
-c
B
\B\ J A- + B-
If B = 0, then Ar + C = 0 is the vertical line x = — C/A. The distance to (xj, yj is
|Aci + C| |Aci + Byi + C|
d =
-(^
|A| ^M2 + B2
(Note that A and B cannot both be zero.)
The slope of the line Ar + By + C = 0 is —A/B. The equation of the line through ix^^, y^ perpendiailar
to Ac + By + C = 0 is:
y~yi = -^yx - x{\
Ay — Ayi = Bx — Bxj
BX[ — .4y| = Bx — Ay
The point of intersection of these two lines is:
Ac + By = -C =^ A-X + ABy = -AC (1)
Bx - Ay = Bx^ - Ay^ => B-x - .4By = B-x, - .4Byi (2)
(A' + B-)x = -AC + B-Xi - AByj (By adding equations (1) and (2))
-AC + B-x^ - AByi
A- + B^
Ac + By =-C => ABx + B2y=-BC (3)
Bx - Ay = Bx, - Avi^ -AB^ + A-y = -ABX) + A-y, (4)
(A- + B-)y = -BC - ABx^ + A^y, (By adding equations (3) and (4))
-BC - ABxi + Ahi^
y =
A- + s-
— CONTINUED—
Section P.2 Linear Models and Rates of Change 13
89. — CO>mNUED—
(-AC + B-x.- ABv, -BC - ABx.+ A'^.\ . ^.
[ ^2 + gi ' ^2 + g2 j point of intersection
The distance between (.r,, y,) and this point gives us the distance between (x,, y\) and the line Ax + By + C = 0.
d= ,
-AC + B-x, - ABy^ _
A^ + S^
-BC - ABxj + Ay
1 _
A- + B-
-AC - ABy, - A-.t,
A= + B-
-BC - ABXj - Bh\
A~ + B-
-A(C + flvi + AC|)
A- + B'
(A^ + B-)[C + At, + By,Y
{A- + B-y
-B(C + /U) + By,)
/12 + B-
^ lAx, + gy, + C\
91. For simplicity, let the vertices of the rhombus be (0, 0),
(a, 0), (b, c), and (o + b, c), as shown in the figure. The
slopes of the diagonals are then
and in.
a + b
b - a
Since the sides of the Rhombus are equal, a^ = ir + c^,
and we have
ib.c) (a+b.c)
(0.0) (a.O)
c c c c
iinu - — — - • 7 - -7 ^ - ^ = - 1.
a + b b - a b — a- - c-
Therefore, the diagonals are perpendicular.
93. Consider the figure below in which the four points are
collinear. Since the triangles are similar, the result imme-
diately follows.
yi - yC yi - >'i
"^ -^1 -^2
95. True.
ax + by = c, => V = -"T-t "*" T"
b b
bx — a\ = C-. => V = — .r '
a a
n,, = --
nu= —
1
14 Chapter P Preparation for Calculus
Section P.3 Functions and Their Graphs
1. (a)/(0) = 2(0)-3 = -3
(b)/(-3) = 2(-3)-3 = -9
(c) fib) = 2i - 3
(d) f(x - 1) = 2(.r - 1) - 3 = It - 5
3. (a) g(0) = 3 - 02 = 3
(b) g(V3) = 3 - (^3)' = 3-3 = 0
(c) g(-2) = 3-(-2)2 = 3-4= -1
(d) g(r - 1) = 3 - (f - 1)2 = -f2 + 2t + 2
5. (a) /(O) = cos(2(0)) = cos 0 = 1
(b,/(-f =c.s2-f =
cosi -yl = 0
'"A!) -4(f)) --3
2£= _i
2
Ai Ax Ax -•(,;.
9.
/(x)-/(2) (i/v^rni-i)
X - 2
X - 2
1 - Vx - 1 1 + Vx - 1
2 -X
(x - 2)Vx - 1 1 + Vx- 1 (x - 2)Vx - 1(1 + Vx - 1) Vx - 1(1 + Vx - 1)
X5t2
11. h{x) = - Vx + 3
Domain: x + 3 > 0 => [-3, 00)
Range: (—00, 0]
13. fit) = sec
77t
Domain: all r t^ 4A' + 2, ^' an integer
Range: (— 00, — 1], [1, 00)
15. fix) = -
X
Domain: (— 00, 0). (0, 00)
Range: (-00, 0), (0, 00)
17. fix) =
2r + l,x < 0
2x + 2, X > 0
(a) /(-1) = 2(-1) + 1 = -1
(b) /(O) = 2(0) + 2 = 2
(c) /(2) = 2(2) + 2 = 6
(d) fif- + 1) = 2(r2 + 1) = 2?2 + 4
(Note: t2 + 1 > 0 for all f)
Domain: (—00, 00)
Range: ( — do, 1), [2, 00)
19. fix) =
\x\ + l,x < 1
-X + l,x > 1
(a) /(-3)= |-3| + 1=4
(b) /(1)= -1 + 1=0
(c) /(3) = -3 + 1 = -2
(d) /(fe'+l)= -(^+ 1)+ 1
Domain: (—00, 00)
Range: (- 00, 0] U [1, 00)
-fc2
Section P.3 Functions and Their Graphs 15
21. fix) = 4- X
Domain: (-00,00)
Range: (-00, 00)
23. h(x) = TT^n"
Domain: [1, oc)
Range: [0, ocj
25. fix) = V9 - x'^
Domain: [—3, 3]
Range: [0,3]
27. ^(f) = 2 sin vt
Domain: ( — oo. oc)
Range: [-2,2]
29. jc - y 2 = 0 => y = ± VI
y is not a function of .t. Some vertical lines intersect
the graph twice.
33. x^ + /
•y
±Va^.
y is not a function of .r since there are two values of v for
some X.
31. V is a function of x. Vertical lines intersect the graph
at most once.
35. ^^ = X-- \
■y = ±v.T-
y is not a function of x since there are two values of >■ for
some X.
37. fix) = |x| + \x-2\
If -T < 0, then/(.r) = -x - U - 2) = -2x + 2 = 2(1 - x).
If 0 < .t < 2, then/(A:) = .t - (.t - 2) = 2.
If X > 2, then/(x) = x + (x - 2) = Ir - 2 = 2fx - 1).
Thus,
[2(1 - x), X < 0
fix) =2, 0 < X < 2.
l2(x - 1), X > 2.
39. The function is ^x) = car. Since (1, —2) satisfies the
equation, c = - 2. Thus, gCr) = - 2xr.
41. The fimction is r{x) = c/x. since it must be undefined at
X = 0. Since (1, 32) satisfies the equation, c = 32. Hius.
rix) = 32/x.
43. (a) For each time t. there corresponds a depth d.
(b) Domain: 0 < r < 5
Range: Q < d < 30
12 3 4 5 6
16 Chapter P Preparation for Calculus
47. (a) The graph is shifted
3 units to the left.
(c) The graph is shifted
2 units upward.
(e) The graph is stretched
vertically by a factor of 3.
r-H 1—
2 4
(b) The graph is shifted
1 unit to the right.
(d) The graph is shifted
4 units downward.
(f) The graph is stretched
vertically by a factor
ofi
49. (a) y = v^x + 2
12 3 4
Vertical shift 2 units upward
(b) y = - v^
Reflection about the ;c-axis
(c) y = V7^^
Horizontal shift 2 units to the
right
51. (a) 71(4) = 16°, 7tl5) - 23°
(b) li H(t) = T{t — 1), then the program would turn on (and off) one hour later.
(c) If //(f) = T{t) - 1, then the overall temperature would be reduced 1 degree.
53. fix) = x\ g{x) = v^
(/ ° g)U) = figix)) = f{V^) = i^f = X, x>0
Domain: [0, oo)
(g 'f)(x) = g{f(x)) = g(;c2) = v<? = |;t|
Domain: (— oo, oo)
No. Their domains are different. {f°g) = (g -/) for x > 0.
55. fix) = -, gix) = ^2 - 1
if'g)i^=figix))=fix^-\)
Domain: all j: # ± 1
(.^/)(.)^.(/(.)) = .g) = gf- 1=1-1=^
Domain: all jc ^^ 0
No,/"g#go/.
Section P.3 Functions and Their Graphs 17
57. {A ■> r)it) = A{r{t)) = A(0.6f) = TT(0.6f)2 = 0.367rt2
{A ' r}{t) represents the area of the circle at time t.
59. f{-x) = (-x)H4 - i-xY) = .t2(4 - x^) =/(jc)
Even
61. f{-x) = (-x)cos(-x) = -jfcosjr = -/W
Odd
63. (a) If/ is even, then (f , 4) is on the graph.
65. f(-x) = a^^ti-x)^"^' + • • • + a3{-jc)3 + a,(-x)
= -K + i^""^' + • ■ ■ +a,x^ + a,x]
= -fix)
Odd
67. Let F(;c) = f(x)g{x) where/ and g are even. Then
F{-x) = /(-.r)5(-.r) = /WgW = F (x).
Thus, F(x) is even. Let FU) = f{x)g{x) where/and g are odd. Then
F(-x) =f{-x)g(-x) = [-f(x)l-g(x)] =f(x)g(x) = Fix).
Thus, F(x) is even.
(b) If/ is odd, then (j, —4) is on the graph.
69. fix) = jc' + 1 and gix) = jt^ are even.
fix)gix) = ix^ + l)(j^) = x« + x^ is even.
5
/(x) = a:^ — x is odd and g{x) = x- is even.
/WgW = ix^ - x)(x^) =x^ - x^h odd.
71. (a)
X
length and width
volume V
1
24 - 2(1)
484
2
24 - 2(2)
800
3
24 - 2(3)
972
4
24 - 2(4)
1024
5
24 - 2(5)
980
6
24 - 2(6)
864
(b)
The maximum volume appears to be 1024 cm',
(c) V = x(24 - 2x)- = 4.t(12 - x)-
Domain: 0 < a < 12
Yes, V is a function of .v.
(d) "00
Maximum volume is V = 1024 cm' for box having
dimensions 4 x 16 x 16 cm.
73. False; let/(x) = x^.
Then/(-3) =/(3) = 9, but -3 9^ 3.
75. True, the function is even.
18 Chapter P Preparation for Calculus
Section P.4 Fitting Models to Data
1. Quadratic function
3. Linear function
5. (a), (b)
250--
200- -
H i 1 1 h*-'
3 6 9 12 15
Yes. The cancer mortality increases linearly with
increased exposure to the carcinogenic substance.
(c) If.:i: = 3, then>'== 136.
7. (a) d = 0.066For F = 15. W + 0.1
(b) 125
The model fits well,
(c) If F = 55, then d » 0.066(55) = 3.63 cm.
9. (a) Let x = per capita energy usage (in millions of Btu)
y = per capita gross national product (in thousands)
y = 0.0764a; + 4.9985 = 0.08x + 5.0
r = 0.7052
(b) «
(c) Denmark, Japan, and Canada
(d) Deleting the data for the three countries above,
y = 0.0959X + 1.0539
(r = 0.9202 is much closer to L)
11. (a) y, = 0.0343f3 - 0.3451^2 + 0.8837t + 5.6061
y2 = 0.1095r + 2.0667
y^ = 0.0917? + 0.7917
(b) 15
y'i^y.
*h
N\ _^.r^^^
nn ^
^
■ ■■'■■
For 2002, t = 12 and ^i + jj + jj = 3 1.06 cents/mile
13. (a) y^ = 4.0367r + 28.9644
^2 = -0.0099;' + 0.5488?^ + 0.2399f + 33.1414
(b) ™
>! =4.04r+29.0
^
^
W^
25
>, = -O.OlOl^ + 0.549l^ + 0.24( + 33. 1
(c) The cubic model is better.
(d) ^3 = 0.4297f2 + 0.5994t + 32.9745
(e) The slope represents the average increase per year
in the number of people (in millions) in HMOs.
(f) For 2000, r = 10, and yj = 69.3 million. Oinear)
yj ~ 80.5 million (cubic)
Review Exercises for Chapter P 19
15. (a) y = - 1.81;c3 + 14.58;t2 + 16.39x + 10
(b) 300
(c) Ifx = 4.5, y = 214 horsepower.
17. (a) Yes, > is a function of t. At each time t, there is one
and only one displacement y.
(b) The amplitude is approximately
(2.35 - 1.65)/2 = 0.35.
The period is approximately
2(0.375 - 0.125) = 0.5.
(c) One model is y = 0.35 sin(4Trf) + 2.
(d)
19. Answers will vary.
Review Exercises for Chapter P
1. >- = 2x - 3
;c = 0 ^> >■ = 2(0) - 3 = -3 =* (0, -3) y-intercept
y = 0:^0 = 2x-3=>jt
(5, 0) .r-intercept
3.y =
X - 1
x-2
X = 0:
y = 0=4>0 =
0 - 1 _ ]_
0-2 2
X - 1
0, -jr I y-intercept
V = 1 =» ( 1 , 0) .T-intercept
5. Symmetric with respect to y-axis since
{-xYy - (-x)' + 4y = 0
xry — X- + 4y = 0.
1 1 _i_ 3
7. y = — 2.V + 2
- vt + fi.V = 1
JA- + y = T
11. V = 7 - dt - X-
y = f .r +
Slope:
y-intercept: j
20 Chapter P Preparation for Calculus
13. y = 75 -;c
Domain: (— oo, 5]
15. y = Ax^- 25
Xmin = -5
Xmax = 5
Xscl = 1
Ymin = -30
Ymax = 10
YscI = 5
17.
3a:-
Ay =
8
Ax +
Ay =
20
Ix
=
28
X =
4
y =
1
Point
: (4, 1)
19. You need factors (x + 2) and {x — 2). Multiply by x to obtain origin symmetry
y = x(x + 2)(x - 2)_
= x^ — 4x.
21.
12 3 4 5
Slope
(5/2) - 1 ^ 3/2 _ 3
5 - (3/2) 7/2 7
23.
1 - t
1 - 5
1-0 1 - (-2)
l-r=-^
25. 3,_(_5) = i(;,_o)
y = F ~ 5
2>' - 3x + 10 = 0
27. y - 0 = -f(x - (-3))
y = -3X - 2
3y + 2x + 6 = 0
Review Exercises for Chapter P 21
29. (a) y - 4 = ~(x + 2)
lo
16y - 64 = 7x + 14
0 = 7.ir - \6y + 78
.X 4-0
y = -2x
2x + y = Q
(b) Slope of line is — .
>- - 4 = jU + 2)
3v - 12 = 5j: + 10
0 = 5;t - 3 V + 22
(d) x=-l
x + 2 = 0
31. The slope is -850. V = -850f + 12,500.
V(3) = -850(3) + 12,500 = $9950
33. a: - r = 0
y = ±Vx
Not a function of j: since there are two values of >> for
some X.
35. y = x^-2x
Function of j: since there is one value oiy for each x.
37. fix) = -
(a) /(O) does not exist.
1
(b)
/(I + Ax) -/(I) ^ 1 + Ax 1 ^ 1 - 1 - A.X:
Ai- Ar (1 + At) Ax
-1
1 + Ar
, Ax9t -1,0
39. (a) Domain: 36 - .v- > 0 ^. -6 < x < 6 or [-6, 6]
Range: [0, 6]
(b) Domain: all x =^ 5 or (-oc. 5). (5, ex:)
Range: all >• ^ 0 or (- oo. 0), (0, oc)
(c) Domain: all x or (-oc, oc)
Range: ally or ( — oc, oo)
41. (a) fix) = x^ + c,c = -2, 0, 2
c = 0
(b) /(x) = (x - c)-\ c = -2,0.
—CONTINUED—
22 Chapter P Preparation for Calculus
41. —CONTINUED—
(c) fix) = ix- 2)3 + c,c
-2,0,2
(d)/(;c) = cx\c= -2,0,2
43. (a) Odd powers: fix) = x, gix) = x^,hix) = x^
r
/
Even powers: fix) = x^, gix) = x*, hix) = j:*
-rA-
The graphs of/, g, and h all rise to the right and fall to The graphs of/, g, and h all rise to the left and to the
the left. As the degree increases, the graph rises and right. As the degree increases, the graph rises more
falls more steeply. All three graphs pass through the steeply. All three graphs pass through the points (0, 0),
points(0,0), (1,1), and (-1,-1). (1, 1), and (- 1, 1).
(b) y = x'' will look like hix) = j^, but rise and fall even more steeply.
y = x^ will look like hix) = x^, but rise even more steeply.
45. (a)
(b) Domain: 0 < jc < 12
2jc + 2>' = 24 . , ,
- )- = 12 - X ,. . ;
A = xy = x(12 - x) = I2x - x^
47. (a) 3 (cubic), negative leading coefficient
(b) 4 (quartic), positive leading coefficient
(c) 2 (quadratic), negative leading coefficient
(d) 5, positive leading coefficient
(c) Maximum area is /I = 36. In general, the maximum
area is attained when the rectangle is a square. In this
case, X = 6.
49. (a) Yes, y is a function of t. At each time t, there is one
and only one displacement y.
(b) The amplitude is approximately
(0.25 - (-0.25))/2 = 0.25.
The period is approximately 1.1.
(c) One model is y
(d) 05
^cos(Yjrj»^cos(5.7f)
Problem Solving for Chapter P 23
Problem Solving for Chapter P
1- (a) ;,2 _ 6;c + y2 _ 8>' = 0
U2 ~ 6x + 9) + {y^-8y+ 16) =9 + \6
U - 3)2 + (y - 4)2 = 25
Center: (3,4) Radius: 5
4-0 4
(c) Slope of line from (6, 0) to (3, 4) is 7 = --.
3 — 6 3
Slope of tangent line is -. Hence,
3 3 9
>" — 0 = —(x — 6) => y = T-T - - Tangent line
(b) Slope of line from (0, 0) to (3, 4) is - Slof>e of tangent line
is -^ Hence,
4
(d)
y-0
3
Hx - 0)
y = --X Tangent line
':x = ':x — —
4 4 2
3
7
3 9
::x = —
2 2
X = 3
Intersection: I 3,
3. H(x) =
1 ;t > 0
0 ^ < 0
I I I 1— ^— i 1 1 h-
" " 12 3 4
(a) H(x) - 2
-4 -3 -2 -1
(b) H(x - 2)
4--
3--
2
I-
-2--
-3--
(c) -H{x)
-2--
-3
-4— I 1 I II— I 1 1 — 1-»-
-4-3-2-1 - - -
(d) H(-x)
4- ■
3--
2- ■
^ — I — I — 6 I — f
J _T _T _l II
-4 -3 -2 -1
-2
.3..
12 5 4
(e) WU)
(f) -H(x - 2) + 2
4-
-4 -3 -2 -1
I I II 1 1 1 >-*-:
-2-
-3-
H 1 — 1 h^
-2-
-3--
24 Chapter P Preparation for Calculus
5. (a) X + ly = 100 =* y =
100 -a:
/100-x\ x-
A(x) = xy = x\ ^ 1 = -— + 5Qx
Domain: 0 < x < 100
(b) 1600
7. The length of the trip in the water is V2^ + x^, and the
length of the trip over land is Vl + (3 - x)^. Hence,
the total time is
_ VaTI? , VI + (3 - xY ,
T = r 1 hours.
Maximum of 1250 m- at j: = 50 m, >> = 25 m.
(c) A{x) = -|(x2 - mx)
= -|(x2 - lOOx + 2500) + 1250
= -|U - 50)2 + 1250
A(50) = 1250 m- is the maximum, x = 50 m, y = 25 m.
9-4
9. (a) Slope = = 5. Slope of tangent line is less than 5.
4 - 1
(b) Slope = _ = 3. Slope of tangent line is greater than 3.
4.41 - 4
(c) Slope = -r-j — = 4. 1 . Slope of tangent line is less than 4. 1 .
(d) Slope
2.1 - 2
/(2 + ;/)-/(2)
{2 + h) -2
(2 + hf- A
h
^ Ah + h^
" h .
= A + h,hi-0
(e) Letting h get closer and closer to 0, the slope approaches 4. Hence, the slope at (2, 4) is 4.
11. (a) At X = 1 and x = — 3 the sounds are equal.
/ 2/
(h)
vVT7 Jix - 3)2 + y-
U - 3)2 + y2 = 4(^2 + y2)
3^2 + 3y2 + 6x = 9
^2 + 2x + / = 3
(X + 1)2 + y2 = 4
Circle of radius 2 centered at (— 1, 0)
Problem Solving for Chapter P 25
13. d^d^ =
[U + 1)^- + ylS.x - \Y + f] =
{x + DHx - 1)2 + y\{x + 1)= + (jc - 1)2] + / =
{x^ - 1)2 + y^[2x^ + 2] + / =
x'^ -2x- + \ + 2xY + 2v2 + / =
(x2 + /)2 = 2(;c2 - /)
Let V = 0. Then x* = 2x- =^ x = 0 or x- = 2.
Thus, (0, 0), {V2, O) and (- V2, o) are on the curve.
(->/2.0)
\
ee
/
(0.0)
CHAPTER 1
Limits and Their Properties
Section 1.1 A Preview of Calculus 27
Section 1.2 Finding Limits Graphically and Numerically 27
Section 1.3 Evaluating Limits Analytically 31
Section 1.4 Continuity and One-Sided Limits 37
Section 1.5 Infinite Limits 42
Review Exercises 47
Problem Solving 49
CHAPTER 1
Limits and Their Properties
Section 1.1 A Preview of Calculus
Solutions to Odd-Numbered Exercises
1. Precalculus; (20 ft/sec)( 15 seconds) = 300 feet
3. Calculus required: slope of tangent line at.ic = 2 is rate of
cliange, and equals about 0.16.
5. Precalculus: Area = \bh = \{5){3) = ysq. units
7. Precalculus: Volume = (2)(4)(3) = 24 cubic units
9. (a)
(b) The graphs of yj are approximations to the tangent line to Vi at j: = 1.
(c) The slope is approximately 2. For a better approximation make the list numbers smaller:
{0.2,0.1,0.01,0.001}
11. (a) D^ = V(5 - 1)' + (1 - 5)- = 716 + 16 « 5.66
(b)D, = yrr(f + 7i + (f-fr + 7i + (f-!r + Vi + (i-ir
« 2.693 + 1.302 + 1.083 + 1.031 =6.11
(c) Increase the number of line segments.
Section 1.2 Finding Limits Graphically and Numerically
1.
X
1.9
1.99
1.999
2.001
2.01
2.1
/w
0.3448
0.3344
0.3334
0.3332
0.3322
0.3226
Iim3
x->2 XT — X — l
~ 0.3333 (Actual limit is |.)
X
-0.1
-0.01
-0.001
0.001
0.01
0.1
fix)
0.2911
0.2889
0.2887
0.2887
0.2884
0.2863
lim ^•^"^^ ^ « 0.2887 (Actual limit is 1/(2^1).)
x-fO X
27
28 Chapter 1 Limits and Their Properties
X
2.9
2.99
2.999
3.001
3.01
3.1
fix)
-0.0641
-0.0627
-0.0625
-0.0625
-0.0623
-0.0610
lim ^^/('+jy^-(y^) ^ _o.0625 (Actual limit is -j-,)
7.
X
-0.1
-0.01
-0.001
0.001
0.01
0.1
fix)
0.9983
0.99998
1.0000
1.0000
0.99998
0.9983
lim = 1.0000 (Actual limit is 1.) (Make sure you use radian mode.)
x-»0 X
9. lim (4 - ;c) = 1
x~*3
11. lim/W = lim (4 - jc) = 2
jr-»2 x->2
13. lim
\x- 5\
does not exist. For values of j: to the left of 5, \x — 5|/(j: — 5) equals - 1,
5 X — 5
whereas for values of a- to the right of 5, \x — 5\/ix - 5) equals 1 .
15. lim tan x does not exist since the fimction increases and 17. lim cos(1/j:) does not exist since the function oscillates
decreases without bound as x approaches tt/2.
between — 1 and 1 as x approaches 0.
19. C(r) = 0.75 - 0.50|-(/- 1)1
(a)
(b)
t
3
3.3
3.4
3.5
3.6
3.7
4
C
1.75
2.25
2.25
2.25
2.25
2.25
2.25
lim Cit) =
r->3.5
2.25
(c)
t
2
2.5
2.9
3
3.1
3.5
4
c
1.25
1.75
1.75
1.75
2.25
2.25
2.25
lim Cit) does not exist. The values of C jump from 1.75 to 2.25 at r = 3.
r->3
21. You need to find 5 such that 0 < |x - 1 1 < 5 implies
l/W-i| =
-- 1
X
< 0.1. That is,
-0.1 < -- 1 < 0.1
X
1 - 0.1 < - < 1 + 0.1
X
A 1 ii
10 ^ ;c "^ 10
10 10
So take 5 = —-. Then 0 < |x - 1 1 < 5 implies
1 , 1
"Tf < -'^ - 1 < 9-
Using the first series of equivalent inequalities, you obtain
1
l/W
< e < 0.
'0 , , 10 ,
1 , 1
9>--l>-H-
Section 1 .2 Finding Limits Graphically and Numerically
29
23. lim (3;c + 2) = 8 = Z,
|(3a + 2) - 8| < 0.01
\'ix - 6| < 0.01
3|jt - 2| < 0.01
0 < |.T - 2| < ^ » 0.0033 = 5
Hence, if 0 < |a; - 2| < S
3|x- 2| < 0.01
\3x - 6| < 0.01
|(3.v + 2) - 8| < 0.01
\f{x) - L\< 0.01
0.01
you have
25. lim (jc^ - 3) = 1 = L
x—*2
|U2 - 3) - 1| < 0.01
\x'^ - 4| < O.OI
|(* + 2){x - 2)1 < 0.01
\x + 2| |.r - 2| < 0.01
0.01
' ' k + 2|
If we assume 1 < x < ?>, then 5 = 0.01/5 = 0.002.
Hence, if0< \x - 2\ < S = 0.002, you have
1^ - 2| < 0.002 = y(0.01) < , I ^,(0.01)
|jc + 2||a:- 2| < 0.01
\x"- - 4| < 0.01
\{x^ - 3) - 1| < 0.01
\f{x) - L\ < 0.01
27. lim (x + 3) = 5
jc->2
Given e > 0:
\{x + 3) - 5| < e
|.r -2| < e= 6
Hence, let S = e.
Hence, if 0 < |.r - 2 1 < 5 = e. you have
\x - 2\ < €
\{x + 3) - 5| < 6
\f(x) -L\<e
29. lim i^x- l) =i(-4)-l = -3
J— »-4
Given e > 0:
\{\x- l)-(-3)| < e
k.+
2 < 6
k\x-{-4)\ < e
\x - (-4)1 < 2e
Hence, let 6 = 2e.
Hence, if 0 < |.r - (-4)| < S = 2e, you have
|;c- (-4)1 < 26
lit + 2| < e
\{kx - l) + 3| < 6
l/W -L\<e
31. lim 3 = 3
x-^6
Given e > 0:
|3-3| < 6
0 < e
Hence, any 8 > 0 will work.
Hence, for any 5 > 0, you have
|3-3|<e
l/W -L\<e
33. Iim4/x = 0
.t-.0
Given e > 0: |.i^ - o| < e
\Vx\ <e
\x\ < e^ = 5
Hence, let 6 = e\
Hence for 0 < |.v — 0| < 6 = e\ you have
|.r| < r'
\^-0\ <€
1/U) -L\<e
30 Chapter 1 Limits and Their Properties
35. lim \x - 2| = |(-2) - 2| = 4
Given e > 0:
\\x - 2| - 4| < 6
|-(;c-2) -4| < e (x - 2 < 0)
\-x-2\ = |;«: + 2| = |;c - (-2)| < e
Hence, 8 = e.
Hence forO < |jr - (-2)| < 5 = e, you have
\x + 2\ < e
\-{x + 2)\ <6
|-U-2)-4| < e
||x - 2| - 4| < e (because a: - 20)
l/W - L| < e
37. lim (x^ + 1) = 2
Given e > 0:
\(x^ + 1) - 2| < e
\x2- 1| < 6
\ix + l)(x - 1)1 < e
U - 1 <
|x+l|
If we assume 0 < a: < 2, then 8 = e/3.
Hence for 0 < |jr - 1 1 < 5 = -, you have
II 1 1
1 < 6
|U= + 1) - 2| < e
l/W - 2| < e
^, , Vx + 5 - 3
^(^^- .-4
lim/W = 1
0
s
The domain is [-5, 4) U (4, oo).
The graphing utility does not show the hole at (4, g)
41. f(x) =
x-9
lim /(a;) = 6
JT— ♦Q
The domain is all x > 0 except x = 9. The graphing
utility does not show the hole at (9, 6).
43. lim/(jc) = 25 means that the values of/ approach 25 as jc gets closer and closer to 8.
45. (i) The values of/ approach different
numbers as x approaches c from
different sides of c:
-4 -3 -2 -1
H — I — I — 1-»-
12 3 4
(ii) The values of/ increase with-
out bound as x approaches c:
(iii) The values of/oscillate
between two fixed numbers as
x approaches c:
4-.
3--
, 47. fix) = (1 + xY
/x
lim(l +xy'' = e« 2.71828
x
/w
X
fix)
-0.1
2.867972
0.1
2.593742
-0.01
2.731999
0.01
2.704814
-0.001
2.719642
0.001
2.716942
-0.0001
2.718418
0.0001
2.718146
-0.00001
2.718295
0.00001
2.718268
-0.000001
2.718283
0.000001
2.718280
12 3 4 5
Section 1.3 Evaluating Limits Analytically 31
49. False; /(x) = (sinx)/jcis
undefined when x = 0.
From Exercise 7, we have
hm = 1.
j:->0 X
51. False; let
/w =
/(4) = 10
S3. Answers will vary.
X- - 4x, X ^ 4
la X = 4'
lim/(x) = lim (x^ - 4x) = 0 i= 10
55. If lim/(x) = L, and lim/(x) = L^, then for every e > 0, there exists S, > 0 and 8-, > 0 such that Ix - cl < 5,
.i->c x->c '
|x — c| < So => |/(x) — LjI < 6. Let S equal the smaller of 5, and Sj. Then for |x — c| < 5, we have
|Li - L2I = |L, -/(x) +/(x) - L2I < |L, -/(x)| + |/(x) - L2I < e + e.
H/(xJ-L,| < eand
Therefore, |Li - L2I < 2e. Since e > 0 is arbitrary, it follows that Lj = Z^.
57. lim [fix) - Z,] = 0 means that for every e > 0 there exists S > 0 such that if
0 < Ix - cl < 5,
then
|(/(x) - L) - 0| < 6.
This means the same as |/(x) - L\ < e when
0 < |x - c| < 5.
Thus, lim/(x) = L.
Section 1.3 Evaluating Limits Analytically
\
1
J
(a) lim h(x) = 0
-v— *5
(b) lim li(x) = 6
13 .t->-l
4
(a) lim/Cx) = 0
x—*0
(b) lim f(x) - 0.524
.t— >Tr/3
(=f)
h(x) = X- - 5x
5. lim.T^ = I'' = 16
fix) = X cos X
7. lim (2x - 1) = 2(0) - 1 = - 1
x->0
9. lim (x^ + 3x) = (-3)2 + 3(-3) = 9-9 = 0
11. lim (2x2 + 4x + 1) = 2(-3)2 + 4(-3) + 1 = 18 - 12 + 1 = 7
13. lim - = I
jr-»2X 2
15. lim
X - 3 1-3
.v^ix= + 4 1- + 4 5
17. lim
_5x 5(7) ^ 35 ^ 35
"" ./FT2 ~ 77 + 2 79 3
19. lim v'x + 1 = V 3 + 1 = 2
-V-.3
32 Chapter 1 Limits and Their Properties
21. lim (x + 3)- = (-4 + 3)- = 1
x-»-4
23. (a) liin/(x) = 5-1=4
(b) lim g{x) = 43 = 64
(c) limg(/W) = g(/(l)) = g(4) = 64
x->l
25. (a) lim/(x) =4-1=3
(b) lim g{x) = 73 + 1 = 2
(c) lim g{f{x)) = g(3) = 2
X->1
27. lim sin jc = sin — = 1
x->7r/2 2
_„ ,- "fx 7r2 1
29. lim cos -r- = cos -^r- = — —
x^2 3 3 2
31. lim sec 2j: = sec 0 = 1
jr->0
„- ,. . . Stt 1
33. lim sm X = sm -7- = —
JC— *5ir/6 6 2
35. lim tan — - = tan — —
x-,3 \ 4 4
= -1
37. (a) lim[5gW] = 5 Urn gix) = 5(3) = 15
(b) lim L/U) + g{x)] = lim/U) + lim gW = 2 + 3 = 5
(c) lim lf{x)g{x)] = riim/(x)]riim g{x)] = (2)(3) = 6
x~^c Ix-^c ilx—*c J
(d) lim ^ = ^^^-7T = ^
x-.cgU) hmgU) 3
39. (a) lim [f{x)f = [lim/(x)? = (4)^ = 64
(b) lim v{70c) = /lim/W = v^ = 2
j:->c v x->c
(c) lim [3/W] = 3 lim/(x) = 3(4) = 12
X— >c x—*c
id) lim[/(x)?^^ = riim/(x)f ^ = (4)V2 = 8
x—*c Lj:— >c J
— 2x^ + X
41. /(x) = - 2jc + 1 and g(j:) = agree except at
jc = 0. ^
(a) lLmg(x) = lim/(x) =1
(b) lim g(x) = lim f(x) = 3
X— > — I X— >-l
43. fix) = xix + 1) and gix) = — — - agree except at x = 1.
(a) lim gix) = lim/(x) = 2
x— >i x— >i
(b) lim g(jc) = lim fix) = 0
X—¥—l X^*-l
X- - 1
45. fix) = — — — and gix) = x - I agree except at x
lim /(jc) = lim gix) = —2
x-*—l x-*-l
x^ - 8
47. fix) = — and gix) = x- + 2j: + 4 agree except at
x = 2.
Vim fix) = lim gix) = 12
X— >2 .r— >2
49. lim
X- 5
lim-
X- 5
x.^5 x^ - 25 X-.5 (x + 5)(x - 5)
= lim
1
1
5X + 5 10
CI r x^ + x-6 ,. ix + 3)(x - 2)
51. lim — , — = Iim
x^-i x^-9
-3 ix + 3)ix - 3)
,. x-2 -5 5
= lim = —7 = 7
Jr-»-3 JC — 3 —6 6
Section 1.3 Evaluating Limits Analytically 33
., ,. V.r + 5 - Vs ,. VxTl - J5 V^ + 5 + 75
53. lim = hm • , 1=
'^0 X x->0 X Jx + 5 + vO
= lim
U + 5) - 5
lim
1
1 75
x^Ox{jx + 5+75) -r->o 7;c + 5 + 75 275 10
,. ,. 7jc + 5 - 3 ,. Jx + 5 - 3 jm + 3
55. lim = hm • ,
x-^i X - 4 x^i X - 4 7^ + 5 + 3
U + 5) - 9 ,. 1
x^4 (x - 4)(7.r + 5 + 3) x^4 Jx + 5 + 3 79 + 3 6
1 1
2 - (2 + x)
._ ,. 2+ x 2 ,. 2(2 +;<:) ,. -1 1
57. hm = hm — ^^ — = lim --; r = --
x-*o x .T-*o X x~*o 2(2 + x) 4
._ ,. 2{x + Ax) - 2x ,. 2x + 2Ax - Ix ,. ^ ,
59. hm -^^ —^ = hm 1 = hm 2 = 2
^x->o A.V A.v->o A.r \x->o
^, ,. {x + Ax)- - 2{x + Ax) + 1 - (.t- - 2;c + 1) ,. x^ + 2xAx + (Ax)- - 2v - 2Ajc + 1 - x= + 2x - 1
61. hm : = lim :
Ai->0 Ax A.x->0 Ax
= lim (2x + Ar - 2) = 2x - 2
„ ,. 7x + 2 - 72 -,,.
63. lim « 0.354
Jr->0 X
X
-0.1
-0.01
-0.001
0
0.001
0.01
0.1
fix)
0.358
0.354
0.345
?
0.354
0.353
0.349
A , • „ >■ v^^"+^- 72 ,. 7^rr2- 72 v^m+ v/2
Analytically, hm = hm • , - — —
x^o X x^o X J7T2 + 72
,. X + 2 - 2 ,.
= lim —, — , ^r = hm
1
./2
0 x(7x + 2+72) --^^o Vx + 2 + v'l 2v'2 4
= 0.354
1 1
65. lim
.t->0
2 +x
X
-0.1
-0.01
-0.001
0
0.001
0.01
0.1
fix)
-0.263
-0.251
-0.250
7
-0.250
-0.249
-0.238
^
1 1
Analytically, lim
2 + X 2
x-*0 X
2-(2+.r) 1 ,. -X 1 ,. -1 1
hm — Z7Z ; — ■ ~ = hm —rz r • — = lim ttt r = — -7-
X .«-^o2(2+.v) X .t-K)2(2 + .x) 4
% 2(2 +.v)
34 Chapter 1 Limits and Their Properties
.„ ,. sin a; ,.
67. lim —1 — = lim
.t->o 5jc j->o
sinx
X
%-
».|i4
„ ,. sin 41 - cos a:)
69. hm — T — — - = lim
1_ sin a: 1 - cos x\
2 X X \
= |(1)(0) = 0
_, ,. sin-x
71. lim = lim
;t-»0 X j:-»0
sin a: = (1) sin 0 = 0
„ ,. (1 - cosh)^ ,.
73. hm ^ ; = hm
h->0 h h-tO
1 - cos h
(1 - cosh)\
(0)(0) = 0
_. ,. cos:r ,. . ,
75. hm = lim sin jc = 1
j:->7r/2 cot X x-^Tt/l
__ ,. Sin 3r ,. / sin 3f
77. hm —r — = hm -— —
,_»o It 1^0 \ 3f
i^-nm-i
2 2
79. m
sin3f
t
-0.1
-0.01
-0.001
0
0.001
0.01
0.1
m
2.96
2.9996
3
7
3
2.9996
2.96
sin 3r ,. ,/sin3f\ ,,,,
Analytically, hm = hm 3 — ;— = 3(1) = 3.
■' t^o t 1^0 \ 3t J
■v^jyj V/Y/~v^
The limit appear to equal 3.
81. fix) =
X
-0.1
-0,01
-0,001
0
0.001
0.01
0.1
fix)
-0.099998
-0.01
-0.001
?
0.001
0.01
0.099998
. , • 1, ,• smj:' ,. /sinr^\ „,,,
Analytically, hm = hm j: — j—i = 0(1) = 0.
im^i \/wvw
o, ,. fix + h)- fix) ,. 2ix + h) + 3 - i2x + 3) ,. 2.x + 2/2 + 3 - 2;c - 3 ,. 2h ,
83. hm-^-^^ r — = lim "^ ; ^ = 1™ ; = hm -- = 2
A-»o h h->o h h^o h '!->o h
„^ ,. fix + h)- fix) ,. x + h X ,. 4x-4ix + h) ,. -4 -4
85. hm-"-!^ f — ^-^ = hm r = hm , ',. , = hm , ^ , , = -^
h-M h h->o h /.->o {x + h)xh /i->o (x + h)x x^
87. lim iA- }?■) < Wm fix) < lim (4 + x^)
x->0 x^O-" x-^0
4 < lim/(;c) < 4
x~*0
Therefore, lim/(j:) = 4.
x-tO
89. fix) = X cos X
lim ix cos jc) = 0
x->0
Section 1.3 Evaluating Limits Analytically 35
91. fix) = \x\ sin X
lim \x\ sinx = 0
;r— »0
93. f(x} = JT sin -
lim I ;c sin - 1 = 0
Ar-»0 V xl
95. We say that two functions /and g agree at all but one
point (on an open interval) if f(x) = g(x) for all jc in the
interval except for x = c, where c is in the interval.
97. An indeterminant form is obtained when evaluating a limit
using direct substitution produces a meaningless fractional
expression such as 0/0. That is,
Jr-.c g(x)
for which lim/(jr) = lim g(x) = 0
99. fix) = X, gix) = sin x, hix) =
S h
^^
'Y
^
When you are "close to" 0 the magnitude of/ is
approximately equal to the magnitude of g.
Thus, |g|/|/| ~ 1 when x is "close to" 0.
101. sit) = -16f2 + 1000
,. si5) - sit) ,. 600 - (-16r= + 1000) ,. 16(r + 5)(/ - 5) ,. ,^, ,^,
hm ^ — = lim —1 ' = lim — ^ — , ' , — - = lim - 16(r + 5) =
t^5 5 - t '->5 5 - r ;->5 -(r - 5) f->5
Speed = 160 ft/sec
103. j(f) = -4.9r2 + 150
,. j(3) - sit) ,. -4.9(3-) + 150 - (-4.9r + 150) ,. -4.9(9 - r)
lim — = lim = lim
1^3 3 - t /-:3 3 - r i->3 3 - f
= lim ""^-^^^ ~ '^^^ '^ '' = lim -4.9(3 + r) = -29.4 m/sec
I-»3 3 - r Jr->3
160 ft/sec.
105. Let/(.r) = l/.v and gix) = - l/.t. lim /(a) and lim ^(.t) do not exist.
.r-*0 -v— +0
lim [fix) + gix)] = lim
.r— >0 .v^O
.V \ X
= lim [0] = 0
.V-.0
107. Given fix) = b. show that for every e > 0 there exists a 5 > 0 such that |/(.v) - b\ < 6 whenever |.v - c\ < 5. Since
|/(.v) - b\ = \b - b\ = 0 < efor any e > 0. then any value of 5 > 0 will work.
109. If fo = 0, then the property is true because both sides are equal to 0. If fc ^ 0. let e > 0 be given. Since lim/(.v) = L.
there exists 5 > 0 such that |/(.t) - L\ < e/\b\ whenever 0 < \x - c\ < S. Hence, wherever 0 < |.v - c-| < 5.
we have
\b\\fix) - L\ < e or \bfix) - bL\ < e
which implies that lim [Z:'/(.v)] = bL.
36 Chapter 1 Limits and Their Properties
111. -M|/(x)| < f{x)g{x) < M\f(x) I
lim(-M|/W|) < limf(x)gix) < lim(M|/W|)
x^c x—*c x—*c
-M(0) < lim f{x)g{x) < M(0)
x—>c
0 < lim f{x)g{x) < 0
x—*c
Therefore, lim f{x)g{x) = 0.
M=_,
113. False. As x approaches 0 from the left, -•— '^ = — 1.
X
I
I
115. True.
117. False. The limit does not exist.
119. Let
fix)
[ 4, if.? > 0
[-4. if j: < 0
lim \f{x)\ = lim4 = 4.
j-»o '■' ' .x->0
lim/(x) does not exist since for a: < 0,/(x) =
l->0
-4andforx > 0,/(x) = 4.
121. f(x) =
g{x) =
0, if x is rational
1, if jc is irrational
0, if x is rational
X, if X is irrational
lim/(x) does not exist.
x—*Q
No matter how "close to" 0 x is, there are still an infinite number of rational and irrational numbers so that lim/(x) does not
exist.
lim g{x) = 0.
x—*0
When X is "close to" 0, both parts of the function are "close to" 0.
123. (a) lim
1 — cos X
1 - cosx 1 + cosx
J-»0 XT
lim ,
jr-»0 jr
1 + cos X
(b) Thus,
1 - cosx 1
1 — cos X ~ —x^
. 1 - cos^x
j:->OX^(1 + COSx)
sin'x 1
■»o jt 1 + cosx
COS X ~ I - -x^ for X = 0.
(c) cos(O.l) = 1 - |(0.1)2 = 0.995
= <=i
(d) cos(O.l) = 0.9950, which agrees with part (c).
Section 1.4 Continuity and One-Sided Limits 37
Section 1.4 Continuity and One-Sided Limits
1. (a) lim fix) = 1
(b) lim fix) = 1
j:->3
(c) lim/W = 1
The function is continuous at
x = 3.
3. (a) lim fix) = 0
(b) lim /W = 0
(c) lim/U) = 0
The function is NOT continuous at
jc = 3.
5. (a) lim fix) = 2
(b) lim /(j:) = -2
X— #4
(c) lim /U) does not exist
J— »4
The function is NOT continuous
at j; = 4.
_ ,. X - 5 1
7. lim ^i -— = lim
x^s- r= - 25
5- ;c + 5 10
without bound as .r
does not exist because
7F^^
grows
11.
,. \x\ ,. -;
lim ■■— ^ = lun —
x->0- X x->0- X
1 1
13.
X + \x X
Ax^O" Ajc
X - ix + A.r) 1
lim — , . . , • -:— = lim
-A.t
1
Ai-»o- A'tr + A.r) zir Ax^o~ .t(.x + Ajc) Lx
= lim -7 — — TT
1
1
a:U + 0)
X + 2 5
15. lim fix) = lim ^ — = -
jc-»3 j:->3 Z Z
17. lim /(.r) = lim (x + 1) = 2
.r— ^1^ x-^l"*"
lim/(.r) = limj.x^ + 1) = 2
lim/W = 2
jr— >1
19. lim cot X does not exist since
X—*TT
lim cot x and lim cot x do not exist.
21. lim (3W - 5) = 3(3) -5 = 4
-t— »4
(H = 3 for 3 < -r < 4)
23.
lim (2 -
x->3
\-
xfj does not exist
because
lim (2 -
-r-*3
-i-
-.^1) =
- 2 -
- (-3) =
5
and
lim (2 -
x—*y
-1-
-.rl) ■
- 2 ■
- (-4) =
= 6.
29. gix) = V25 - .t- is continuous
on [-5, 5],
25. fix)
X- - 4
has discontinuities at .t = — 2 and
.r = 2 since/(-2) and/(2) are not
defined.
31. lim fix) = 3 = lim f(x).
x->0- .t->0*
/is continuous on [— 1, 4].
27./W = H + ..
has discontinuities at each integer
k since lim f[x) ~ lim /(jr).
x—^k' x—*k'
33. /(.t) = .t- - It + 1 is continuous
for all real x.
38 Chapter 1 Limits and Their Properties
35. f(x) = 3x — cos X is continuous for all real x.
37. fix) = -^ is not continuous at x = 0, 1 . Since
I
, ,■ for jc =^ 0, ;: = 0 is a removable
X'^ — X X — \
discontinuity, whereas x = 1 is a nonremovable
discontinuity.
39. fix)
x^+ 1
is continuous for all real x.
41. fix)
x + 2
ix + 2)ix - 5)
has a nonremovable discontinuity at j: = 5 since lim/(x)
does not exist, and has a removable discontinuity at
X = —2 since
lim fix) = lim
\x + 2|
43. fix) = -' — — r"- has a nonremovable discontinuity at j: = - 2 since lim fix) does not exist.
45. fix) =
X, X < \
X?-, X > \
has a possible discontinuity at x = 1.
1. fix) = 1
lim /W = lim X = r
lhn/(jc) = limx^ = 1.
3. /(I) = lim/(jc)
lim/W = 1
/is continuous alx = 1, therefore, /is continuous for all real x.
, ^ r^ + 1, X <2
47. /U) = \ ^ has a possible discontinuity at x = 2.
l3 - X, X > 2
1. /(2) =-+1=2
lim/(.r) = lim (^+ 1) = 2]
2_ Mim/(x) does not exist.
lim fix) = lim (3 - x) = 1 J"""*
Therefore, /has a nonremovable discontinuity atx = 2.
49./(x) = r"4-
tan
^. U < 1
tan^. -1 <x < 1
1. /(- 1) = - 1
2. lim fix) = - 1
X— ♦ — 1
3./(-l)= lim/(x)
X—*~l
/(I) = 1
lim/(x) = 1
/(I) = lim/(x)
x—*l
X < — 1 or X > 1
has possible discontinuities atx= — l,x= 1.
/is continuous at x = ±1, therefore, /is continuous for all real x.
Section 1.4 Continuity and One-Sided Limits 39
51. f(x) = CSC 2x has nonremovable discontinuities at integer
multiples of ir/2.
53. fix) = [jc - ll has nonremovable discontinuities at each
integer k.
55. lim f(x) = 0
x—*0*
lim fix) = 0
/is not continuous at x = -2. -s
\
/
c/
57. /(2) = 8
Find a so that lim or = 8
^ = ^ = 2.
59. Find a and b such that lim (ax + ^) =
a - fo= -2
(+)3a + fe = -2
4a = -4
a= -I
b= 2 + (-1) = I
-a + b = 2 and lim (ox + fc) = 3a +
2, .t < - 1
/U) = |-;c + 1, -1 < X < 3
1-2, jc > 3
-2.
61. /(gU)) = ix - \Y
Continuous for all real x.
63. figix))
1
U- + 5) - 6 ;c- - 1
Nonremovable discontinuities at ;c = ± 1
65. y = ix\-x
Nonremovable discontinuity at each integer
^;mx
67. fix) =
2x - 4, .r < 3
Nonremovable discontinuity at .v = 3
5
I
/
/
/
69. fix)
X- + 1
Continuous on (— oo, oo)
71. /U) = sec^
Continuous on:
. . .,(-6. -2), (-2. 2), (2. 6), (6. 10),
73. fix) =
3
75. fix] = -^xf' - .v-^ + 3 is continuous on [l. 2].
/(I) = ^ and /(2) = -4. By the Intermediate Value
Theorem, /(c) = 0 for at least one value of c between
1 and 2.
The graph appears to be continuous on the interval
[-4, 4]. Since /(O) is not defined, we know that/has
a discontinuity at .v = 0. This discontinuity is removable
so it does not show up on the graph.
40
Chapter 1 Limits and Their Properties
77. f{x) = jc^ — 2 — cos X is continuous on [0, it].
/(O) = -3 and /(it) = 77^ - 1 > 0. By the Intermediate
Value Theorem, /(c) = 0 for the least one value of c
between 0 and -tt.
79. f{x)=x^ + X- \
f(x) is continuous on [0, 1].
/(O) = - 1 and/(l) = 1
By the Intermediate Value Theorem, f{x) = 0 for at least
one value of c between 0 and 1 . Using a graphing utility,
we find that j: = 0.6823.
81. git) = 2 cos r - 3r
g is continuous on [0, 1].
g{0) = 2 > Oandg(l)«
1.9 < 0.
By the Intermediate Value Theorem, g(t) = 0 for at least
one value c between 0 and 1. Using a graphing utility, we
fmd that t ^ 0.5636.
83. fix) = x^ + X- I
f is continuous on [0, 5].
/(O) = - 1 and/(5) = 29
-1 < 11 < 29
The Intermediate Value Theorem applies.
a:2 + jc - 1 = 11
x^ + X - 12 = 0
ix + 4)(;c - 3) = 0
x=—4oTX = 3
c = 3 {x = —4 is not in the interval.)
Thus,/(3) =11.
85. fix) = x^-x^ + x-2
/is continuous on [0, 3].
/(0) = -2and/(3) = 19 ,, ■
-2 < 4 < 19
The Intermediate Value Theorem applies.
x^-x^ + x-2 = 4
x^ + x
0
ix - 2)ix^ + ;c + 3) = 0
X = 2
(x^ + X + 3 has no real solution.)
c = 2
Thus,/(2) = 4.
87. (a) The limit does not exist at x = c.
(b) The function is not defined at x = c.
(c) The limit exists at -"^ ~ i^' but it is not equal to the
value of the function di^ ^ ^■
(d) The limit does not exist at x = c.
89.
The function is not continuous at x = 3 because
lim /(x) = 1 # 0 = lim fix).
91. The functions agree for integer values of x:
gix) = 3 - I-xl = 3 - (-x) = 3 + X "
fix) = 3 + W = 3 + X
for X an integer
However, for non-integer values of x, the functions
differ by 1 .
fix) = 3 + M = g(x) - 1 = 2 - I-xl.
For example, /(j) = 3 + 0 = 3, g{^ = 3 - (- 1)
Section 1.4 Continuity and One-Sided Limits 41
93. Mf) = 25 2
f + 2
- t
t
0
1
1.8
2
3
3.8
N{t)
50
25
5
50
25
5
Discontinuous at every positive even integer.Tlie
company replenishes its inventory every two months.
; 4 6 8 10 12
Tune (in months)
95. Let V = 3 TTr^ be the volume of a sphere of radius r.
V(l) =|7r«4.19
V(5) =3tt(53)-523.6
Since 4.19 < 275 < 523.6, the Intermediate Value Theorem implies that there is at least one value r between 1 and 5 such
that V{r) = 275. (In fact, r « 4.0341.)
97. Let c be any real number. Then \imf(x} does not exist since there are both rational and
x—*c
irrational numbers arbitrarily close to c. Therefore, /is not continuous at c.
-1, if.v < 0
99. sgn(.T) = 1 0, if a: = 0
1, ifx > 0
(a) lim sgnU) = - 1
(b) lim^ sgn(jc) = 1
(c) lim sgn(jr) does not exist.
x—^Q
4--
3--
2
1*
-4 -3-2-1
H 1 1 — 1-»-
12 3 4
101. True; iff(x) = g(x), x i- c, then lim/U) = lim 2,{x) and
X~*C .1— .c
at least one of these limits (if they exist) does not equal
the corresponding function s^x = c.
103. False: /(I) is not defmed and lim/(.r) does not exist.
X— »I
105. (a) f{x) =
(0 0 < .X < fc
\b b < x <2b
NOT continuous a.tx = b.
(b) g{x) =
0 < X < b
b -- b <x <2b
7
Continuous on [0. 2b].
42 Chapter 1 Limits and Their Properties
107. f{x) = ^''^'^ £, c > 0
Domain: x + (P->Q=^x> -c^andxT^o, [-c^, 0) U (0, oo)
•Jx + c~ — c ^x + c~ + c
,. Jx + c - c
lim = iim ■
x->0 X x^O X
JW? + c
lim •
U + c^) -c^ .. 1
•'-»'' \jx + c^ + c] '^0 Va: + c^ + c Ic
Define /(O) = l/(2c) to make /continuous at jc = 0.
109. h[x) = x\x\
h has nonremovable discontinuities at
x = ±l,±2,±3, ... .
"^
Section 1.5 InOnite Limits
1.
lim 2
, ,. TTX
3. lim tan — -
j:->-2* 4
lim 2
-»-2-
x^ -A
TTX
liin tan — - = oo
;c-»-2- 4
5. /W =
X
-3.5
-3.1
-3.01
-3.001
-2.999
-2.99
-2.9
-2.5
fix)
0.308
1.639
16.64
166.6
- 166.7
- 16.69
-1.695
-0.364
lim f(x) = oo
j:->-3
lim /U) = -oo
j:-*— 3
7. /U)
;c2-9
x
-3.5
-3.1
-3.01
-3.001
-2.999
-2.99
-2.9
-2.5
fix)
3.769
15.75
150.8
1501
-1499
- 149.3
- 14.25
-2.273
Vim fix) = oo
lim fix) = -oo
x—*-3*
Section 1.5 Infinite Limits 43
9. lim ^ = oo = lim -^
jr-»0* X^ i-»0~ X^
Therefore, at = 0 is a vertical asymptote.
"•ii.'?^(x-2)U+l) = °°
Therefore, j: = 2 is a vertical asymptote.
.-l'?r(;c-2)U+l) = °°
;t2-2
lim
-i-U-2)U+ 1)
Therefore, x = — 1 is a vertical asymptote.
x^ X-
13. lim -:; 7 = oo and lim
Jr->-2- X' — 4
' Jr-U"-2+ X^ - 4
Therefore, x = — 2 is a vertical asymptote.
T-2 r2
lim
-oo and lim
x->2- x^ — 4 """ -t'-^r jr2 - 4
Therefore, jt = 2 is a vertical asymptote.
15. No vertical asymptote since the denominator is never zero.
17. fix) = tan 2x = — has vertical asymptotes at
cos It
{2n + \)tt tt n-TT
X = = — + — , n any mteger.
19. lim 1 - -
r->0* V t-
-OO = lim 1 :;
1^0- V r-
Therefore, r = 0 is a vertical asymptote.
21. lim
x^-2- (x + 2)(x - 1)
lim 7 TTT TT = -oo
x-»-2- (x + 2)(x - 1)
Therefore, x = - 2 is a vertical asymptote.
lim -; — 7 -r = oo
x-^r (x + 2)(x - 1)
lim '. TT-, rr = — oc
x-.r (x + 2)(x - 1)
Therefore, x = 1 is a vertical asymptote.
23. Ax) =
A^ + 1 _ (.T-t- DU^ --r + 1)
X + 1
X + 1
has no vertical asymptote since
lim fix) = lim ix- - .r + 1) = 3
X— *— 1 J—*- I
25. fix.
5)(.r + 3) _ x + 3
X* 5
ix - 5)(a:- +1) X- + V
No vertical asymptotes. The graph has a hole at x = 5.
27. sit) = — — has vertical asymptotes al t = mr. n
smt
a nonzero integer. There is no \ertical asymptote at
t = 0 since
lim^— = 1.
t->o sm t
44 Chapter 1 Limits and Their Properties
29. lim i-— f- = Urn [x - 1)
jr-»-l X + \ Jr-»-l
31. lim — —r = oo
;t-»-l+ X + 1
;t^+ 1
lim
\- x+ \
Vertical asymptote at
x= -1
Removable discontinuity at a: = — 1
33. lim r-
x-,2* X - 1
35. lim
x^-i- (x - i)(x + 3)
,_ ,. x^ + 2x- ^ ,. X- \ 4
37. lim -^5— — = lim = 7
x-^-r XT + X — it j:-»-3 X — I 5
A — X X 1
x^l (x + l)(x - 1) x-H X^ + I 2
41. lim 1 + - = -00
43. lim -: — = 00
jr-»o+ sin X
45. lim = lim (v^sin;c) = 0
.r-»irCSCJ: x^-ir
47. lim X seclTTx) = 00 and lim x sec(irx)
^-^(1/2)- xMU2r
Therefore, lim x secfTTj:) does not exist.
x-*(l/2)
49. f(x) = -^^-j-
51. fix) =
x^- 25
lim f(x) = lim = 00
lim fix) = -00
x-*5
J
L
r
"^i
53. A limit in which fix) increases or decreases without
bound as x approaches c is called an infinite limit. 00 is
not a number. Rather, the symbol
lim fix) = 00
x-*c
says how the limit fails to exist.
55. One answer is fix) =
x-3
X- 3
ix - 6)(;c + 2) x^--4x- 12'
57.
59. 5 = -; , 0 < Irl < 1. Assume k i= 0.
1 - r
lim S = lim
r->r r->i- 1 — r
= 00 (or — 00 if fc < 0)
Section 1.5 Infinite Limits 45
^, ^ 52&X „
61. C = -— , 0 < X < 100
100 - X
(a) C(25) = $176 million
(b) C(50) = $528 million
(c) C(75) = $1584 million
(d) lim
528
-.100- 100 - ;c
OD Thus, it is not possible.
2(7) 7
63. (a) r = , ' = — ft/sec
(b) r-
(c) lim
7625 - 49 12
2(15) 3
V625 - 225 2
ft/sec
--^^25- V625 - x'-
65. (a)
X
1
0.5
0.2
0.1
0.01
0.001
0.0001
fix)
0.1585
0.041 1
0.0067
0.0017
= 0
= 0
= 0
X - sm.t
lim —■ = 0
j:->0* X
(b)
X
1
0.5
0.2
0.1
0.01
0.001
0.0001
fix)
0.1585
0.0823
0.0333
0.0167
0.0017
«0
«0
;c - smx
lim :; = 0
(c)
X
1
0.5
0.2
0.1
0.01
0.001
0.0001
fix)
0.1585
0.1646
0.1663
0.1666
0.1667
0.1667
0.1667
0.25
1.5
lim - — ^ = 0.1167 (1/6)
x->0* XT
(d)
X
1
0.5
0.2
0.1
0.01
0.001
0.0001
fix)
0.1585
0.3292
0.8317
1.6658
16.67
166.7
1667.0
X - sin.T
Iim ; = oo
.1^0* .V-*
^ , ,. .V - sin.v
For n > i, lim ;; = oo.
x-^O* X"
46
Chapter 1 Limits and Their Properties
67. (a) Because the circumference of the motor is
half that of the saw arbor, the saw makes
1700/2 = 850 revolutions per minute.
(c) 2(20 cot </)) + 2(10 cot (^): straight sections.
The angle subtended in each circle is
277
2(f-^
77+ 2</).
Thus, the length of the belt around the pulleys is
20(77 + 24>) + 10(77 + 24>) = 30(77 + 24>).
Total length = 60 cot ^ + 30(77 + 2</))
Domain: ( 0, —
69. False; for instance, let
f(x) = r or
six)
X - 1
x^+ 1
(b) The direction of rotation is reversed,
(d)
(e) 450
<!>
0.3
0.6
0.9
1.2
1.5
L
306.2
217.9
195.9
189.6
188.5
(0 lim L = 6077 == 188.5
«^(t/2)-
(All the belts are around pulleys.)
(g)
lim L =
00
71.
False; let
f{x) =
§
;c # 0
;c = 0.
The graph of /has a vertical asymptote at x = 0, but
m = 3.
73. Given lim/(j:) = cxd and lim g{x) = L:
x—*c x~*c
(2) Product:
If L > 0, then for e = L/2 > 0 there exists 5; > 0 such that \g{x) - L\ < L/2 whenever 0 < |a: - c| < S,. Thus,
L/2 < g{x) < 3L/2. Since lim/(j:) = 00 then forM > 0, there exists 5, > 0 such that /(a:) > M(2/L) whenever
|jc - c| < S;. Let 5 be the smaller of 5i and Sj- Then for 0 < |a: - c| < S, we have/(;t)g(x) > M(2/L){L/2) = M.
Therefore Vim f(x)g{x) = 00. The proof is similar for L < 0.
(3) Quotient: Let e > 0 be given.
There exists S, > 0 such that/(jr) > 3L/2e whenever 0 < \x — c\ < S^ and there exists 63 > 0 such that \g{x) ~ L\ <
L/2 whenever 0 < |x - c| < S,. This inequality gives us L/2 < g{x) < 3L/2. Let S be the smaller of 5, and Sj. Then
for 0 < \x — c\ < d, v/e have
g(x)
Ax)
3L/2
3L/2e
e(x)
Therefore, lim 7H = 0.
x-*cf(x)
75. Given lim -7-, = 0.
x^cf(x)
Suppose lim/(j:) exists and equals L. Then,
x-*c
liml
1
1
x->cf{x) lim/(;c) L
This is not possible. Thus, Umf(x) does not exist.
Review Exercises for Chapter 1 47
Review Exercises for Chapter 1
1. Calculus required. Using a graphing utility, you can estimate the length to be 8.3.
Or, the length is slightly longer than the distance between the two points, 8.25.
3.
X
-0.1
-0.01
-0.001
0.001
0.01
0.1
fix)
-0.26
-0.25
-0.250
-0.2499
-0.249
-0.24
lim/W « -0.25
5. h{x)
)r - 2x
(a) \\mh{x) = -2
x-»0
(b) lim h{x) = -3
7. lim (3 - a:) = 3 - 1 = 2
Let e > 0 be given. Choose 6 = e. Then for
0< |x-l| <5=6, you have
\x-\\<e
\\-x\<e
1(3 -x)-2\<e
\f(x) -L\<e
9. lim (x- - 3) = 1
Let e > 0 be given. We need |r= - 3 - 1 1 < e => |.t- - 4| = \{x - 2)(a: + 2)| < e ^ |.r - 2| <
Assuming, 1 < .v < 3, you can choose 6 = e/5. Hence, for 0 < |.x — 2| < S = e/5 you have
x + 2
X- 2 <T <
1
5 \x + 2\
\x- 2\\x + 2\ < e
\x- - 4| < e
l(x^-3)- 1| < e
l/W - L\<e
11. lim Vm = V4 + 2 = V6 = 2.45
<-»4
13. lim "5 7 = lim = —-
r->-2 r — 4 r->-2 t - 2 4
Jx - 2
15. lim — '■ = lim
Vi-2
-.4 X - 4
■4(v3^-2)(^ + 2)
lim ^ = — F = -
x-*'i Vx + 2 V4 + 2 4
„ ,. [l/(.v + D] - 1 ,. 1 - Ct + 1)
17. lim ^= = lim — : — . .,
.t->0 X Jr->0 x(x + 1)
1
= lim
i^>ox + 1
= -I
,n r ^ + 125 ,. (.V + 5)(.r - 5.t + 25)
19. lim — r- = lim
-5 X + 5
X + 5
,, ,. 1 - COS.V ,. / X \(l -COS.tA ,,Mr,\ rv
21. hm : = lim ^ — = (1)(0) = 0
x-»o sm.v x-'0\,sina:/\ .t /
lim (r - 5x + 25)
= 75
48 Chapter 1 Limits and Their Properties
,, ,. sin[(7r/6) + Ajc] - (1/2) ,. sin(Tr/6) cos A;c 4- cos(7r/6) sin Ax - (1/2)
23. lim : = lim :
Ax-»0 Ar Ax->0 A.t
1 (cos Ajc- 1) ,. V^ sin Ax
= lim - • -. H lun —T- • — : —
Aj-»o 2 Ax Ax-*o 2 Ax
25. lim[/(x)-g(.x)] = (-|)(f)=-|
27./(x) = ^-j
(a)
j;
1.1
1.01
1.001
1.0001
fix)
0.5680
0.5764
0.5773
0.5773
lim + 1 V3 ^ Q^^^ (Actual limit is ^3/3.^
J— ♦l^ ^ — 1
,, ,. J2x +1-73 ,. V2x + 1 - 73 72x +1 + 73
(c) lim = lim ; • , p
j:-i* .X - 1 ^-»i* x-1 72x +1 + 73
(2x + 1) - 3
~ x-^v (x - l)(72x + 1 + v^)
2
= hm — , =
^^'" 72x+ 1 + 73
2 1 _^
2VS 73 3
(b)
-_ ,. s{a) - s{t) ,. (-4.9(4)- + 200) - (-4.9r + 200)
29. lim = lim
r-»o a — t »->4 4 — t
,. 4.9(r - 4)(f + 4)
= lim :
r-*4 A — t
= lim -4.9(r + 4) = -39.2 m/sec
I->4
31. lim
\x- 31
j:^3- X: — 3
lim^^=-l
Jr^3- X — 5
33. lim/(x) = 0
x-*2
35. lim /!(f) does not exist because lim h{i) = 1 + 1=2 and
r— >I r— »1"
lim h(t) = i(l + 1) = 1.
37. fix) = Ix + 31
lim [x + 31 = /c + 3 where ^ is an integer.
lim |x + 3]1 = /c + 2 where A: is an integer.
Nonremovable discontinuity at each integer k
Continuous on {k, k + I) for all integers k
39. fix) =
3x- - x-2 (3x + 2)(x: - 1)
1
1
lim/(x:) = lim (3.x + 2) = 5
J— » I Jr— > 1
Removable discontinuity at x = 1
Continuous on (— oo, 1) U (1, oo)
41. fix)
lim
1
ix - 2)2
1
x^2 (;c - 2)2
Nonremovable discontinuity at x = 2
Continuous on (-co, 2) U (2, oo)
43. fix) =
X + 1
lim /(x) = — oo
JC-+1
lim /(x:) = oo
jif-*i*
Nonremovable discontinuity at x: = — 1
Continuous on (-oo, — 1) U (- 1, oo)
Problem Solving for Chapter 1 49
45. f{x) = CSC —
Nonremovable discontinuities at each even integer.
Continuous on
(2k, 2k +2)
for all integers k.
49. /is continuous on [l, 2]. /(I) = - 1 < 0 and
/(2) = 13 > 0. Therefore by the Intermediate Value
Theorem, there is at least one "alue c in (1, 2) such
that 2c5 - 3 = 0.
47. /(2) = 5
Find c so that lim (ex + 6) = 5.
c(2) + 6 = 5
2c = -1
1
c=--
51. fix) = ^^ ={x + 2)
\ X -2
Lk-2|
(a) lim/U) = -4
jr-»2
(b) lim^/W = 4
(c) lim/(jc) does not exist.
53. gW = 1 + -
Vertical asymptote at x = 0
2x^ + X+ \
57. lim
X + 2
55. /(.r
(x - 10)=
Vertical asymptote at x = 10
59. lim 4-^-
1
J-*-!* X-
_ ,. .r- + 2,t + 1
61. lim ;
j:-»r j; - 1
63. lim [x J
sin 4x
65. lim — - — = lim
.r->0* 5x X-.0*
4/ sin 4a:
5V 4x
„ ,. csclv ,. 1
67. lim = lim — ^ — — = oo
.v->o* X X ->o* X Sin Zr
69.C = |^,0.0<100
100 — p
(a) C(15) == $14,117.65 (b) C(50) = $80,000
(c) C(90) = $720,000
(d) lim tt:: = oo
p->ioo' 100 - p
Problem Solving for Chapter 1
1. (a) Perimeter APAO = Jx- + (v - 1)= + Jx- + y- + 1 (b) r(x)
Jx- + (.V-2 - 1): + Jx- + X* + 1
Perimeter APBO = V(.t - 1)- +r + n/^+T + 1
V(a- - IP + A^ + 7^2 + A-^ + 1
(c) hm r(x) = . , „ , . = T = 1
.v->o+ 1+0+1 2
x- + ix' - \)- + ^x~ + .1-' + 1
Jix - ly- + x" + ^'x^ + .x-* + 1
.X
4
2
1
0.1
0.01
Perimeter AP.-^O
33.02
9.08
3.41
2.10
2.01
Perimeter APBO
33.77
9.60
3.41
2.00
2.00
r{x)
0.98
0.95 1
1.05
1.005
50 Chapter 1 Limits and Their Properties
3. (a) There are 6 triangles, each with a central angle of
60° = 77-/3. Hence,
Area hexagon = 6
■bh
2
3V3
2(1) Sin -
- 2.598.
h = sme
Error: tt r — = 0.5435.
(b) There are n triangles, each with central angle of
6 = iTr/n. Hence,
An = n
bh
1,,, . Itt] nsmilv/n
:r(l sin — =- — ^r-^
2 « J 2
n
6
12
24
48
96
An
2.598
3
3.106
3.133
3.139
(c)
(d) As n gets larger and larger, In/n approaches 0.
Letting x = 27r/n,
sin(2ir/«) sin(2Tr/n) sinj:
An = — —. = , , , 77 = 77
2/n (277/m) j:
which approaches (1)77 = 77.
5. (a) Slope = —
12
(b) Slope of tangent line is
y+l2 = —{x- 5)
169
12'
y = — JT - — Tangent line
(c) Q=(x,y) = {x,^l69-x^)
- V169 - jc2 + 12
X — 5
,. 12 - V169 - x^ 12 + V169 - x^
(d) lim m^ = lim ;: • ,
-t^s ^ ;t^5 X - 5 12 + Vl69 - ;c2
^ 144 - (169 - x^)
" ^™ (;c - 5)(l2 + ^169"^^)
= lim
x^ - 25
= lim
^5 (x - 5)(l2 + V169 - x^)
ix + 5)
5 12 + Vl69 - x^
10 5
12 + 12 12
This is the same slope as part (b).
7. (a) 3 + ;c'/3 > 0
a:'/3 > -3
;c > -27
Domain: x > —27,xi= 1
(b)
(c) lim J{x) =
V3 + (-27)'/3 - 2
-27 - 1
-2 1
-28 14
0.0714
. ,, ,. ., , ,. V3 + -t'/^ - 2 V3 + x'/3 + 2
(d) lim/(x) = lim ; • . ,,:
= lim
3 + ;c'/3 - 4
= lim
1 (x - l^v/sT^iTs + 2)
;c'/3 - 1
^-^1 (xi/3 - l)(;c2/3 + ^1/3 + i)(V3 + x^/i + 2)
1
= lim
^->l (x2/3 + ;t'/3 + 1)(V3 + ;c'/3 + 2)
1
(1 + 1 + 1)(2 + 2) ~ 12
9. (a) lim/W = 3: ^„ ^4
fb) /continuous at 2: g,
(c) lim fix) = 3: g„ g,, g^
X— ♦2
Problem Solving for Chapter J 51
11.
-2--
-3--
13. (a) y
a b
(a) /(1) = I11 + I-]1= 1 +(-l) = 0
/(O) = 0
/(i) = 0 + (-l)=-l
/(-2.7) = -3 + 2 = -1
(b) \\mf(x) =
-1
lim fix) =
Jr->1*
-1
Junjix) =
-1
(c) /is continuous for all real numbers except
X = 0,±l,+2,±3, . . .
(b) (i) lim^ P„ ,{x) = 1
x—*a
(ii) lim P„ ,{x) = 0
(iii) lim P„ ,{x) = 0
X— >o
(iv) lim P^,(x) = 1
x—*b
(c) Pn ^ is continuous for all positive real numbers
except x = a,b.
(d) The area under the graph of u,
and above the j:-axis, is 1.
CHAPTER 2
Differentiation
Section 2.1 The Derivative and the Tangent Line Problem ... 53
Section 2.2 Basic Differentiation Rules and Rates of Change . 60
Section 23 The Product and Quotient Rules and
Higher-Order Derivatives 67
Section 2.4 The Chain Rule 73
Section 2.5 Implicit Differentiation 79
Section 2.6 Related Rates 85
Review Exercises 92
Problem Solving 98
CHAPTER 2
Differentiation
Section 2.1 The Derivative and the Tangent Line Problem
Solutions to Odd-Numbered Exercises
1. (a) m = 0
(b) m = -3
3. (a), (b)
(c).= ^^f^U -!)-/(!)
:(x- l) + 2
l(x - 1) + 2
X + I
12 3 4 5 6
5. fix) = 3 — 2j: is a line. Slope
= lim
AI-.0
/(0 +
Af)-/(0)
Af
= lim
A/->0
3(Ar)
- (Af)' - 0
Af
= lim
M->0
(3-
Ar) = 3
13. f(x) = -5x
fix) = lim
= lim -
A;t-»0
/(.t+ Ax)-/(a:)
zSa
-5U + .A-v) - (-5.t)
A.r
lim — 5 = -5
Al->0
7.
Slope
at (1,-3) =
= lim
Aj-»0
g(i + ^x) -
\x
gi\)
=
= lim
(1 + ^xy- -
A.r
4 -(-3)
=
= lim
1 + 2(A.t) +
Ax
(Ax)2 - 1
=
= lim
Aj:->0
[2 + 2(^x)']
= 2
11.
/(.O
= 3
ru)
= lim f^^
Ai->0
+ A.r)
A.r
-fix)
= lim — —
Ai->0 Ax
3
= lim 0 =
0
15
/i(i) =
= 3.f.
h'is)
= lim -^
+ ^s)
-his)
3+ j(5 +
ls)-{i^
f>)
A5^0
Is
= lim -;—
As-KI Aj
2
"I
53
54 Chapter 2 Dijferentiation
17. f[x) = 2.t2 + X - 1
f(x+ ^x)-f{x)
J \X)
— um
A.r
= lim
A1-.0
[2(;c + ^xf + [x + hx) -
-1]-
[2r + X - 1]
^x
= lim
{Ix" + 4x^x + liH^f +
;c + ^
- l)-
(2^2 +
X- 1)
Aa
= lim
Ax->0
Ax^x + 2(Ax)^ + Ajc ,. ,.
-, — = lim (Ax
Ajc ax-»o
+ 2Ax + 1) =
4x+ 1
19. fix)
= X3-
12x
fix)
= lim
f{x + ^x)-f{x)
Ax
= lim
Ax->0
[(x + AxY - \2(x + Ajc)
1-U'
-12x]
Ajc
= lim
x^ + 3x^Ax + 3a;(A;c)2 +
(AxP-
12x-
12A;c-
- x3 + 12x
Ax
= lim
Ax^O
3;c2Ax + 3;c(A;c)2 + (A;c)3
Ax
- 12Ax
= lim (3x2 + 3^^ + (^)2 _ j2) = 3^2 - 12
Ax->0
21. fix) '
/'(x) = lim
X- 1
/(x + Ax)-/(x)
Ax->0 Ax
1 1
x + Ax-1 X — 1
= lim 1
Ax-»0 Ax
U - 1) - (x + Ax - 1)
~ iJc^o Ax(x + Ax - l)(x - 1)
= r -Ax
iJ™oAx(x + Ax- l)(x- 1)
" iji^o (x + Ax - l)(x - 1)
^ 1
u-ip
23. /(x) = Vx+ 1
/'(x) = lim ^<-^y-^(-)
•^ Ax^O Ax
Vx + Ax +T - Vx + 1 _ /Vx + A.r + 1 + Vx + 1\
^^0 Ax VVxTAxTT + Vx + 1/
Ax-
(x + Ax + 1) - (x + 1)
= lim
Ax-*o Ax[Vx + Ax + 1 + Vx + 1]
lim , ^ ,
■^-'0 Vx + Ax + 1 + Vx + 1
1 ^ 1
Vx + 1 + Vx + 1 2Vx + 1
Section 2.1 The Derivative and the Tangent Line Problem 55
25. (a) fix) =x^ + \
fix) = lim
fix + Ax) -fix)
Ax
^ ^.^^^ [jx + Ax)^ +\]-[x'+l]
Ax-»0 AjC
,. 2xA;c + (A.r)^
= lim ;
A.t->0 Ax
= lim (Ix + A.x) = 2x
At (2, 5), the slope of the tangent line is
m = 2(2) = 4. The equation of the tangent line is
y-5
= 4(.t
-2)
>'-5
= 4x
- 8
y
= 4x
- 3.
27.
(a) fix) =
;c3
fix) =
lim
Ax->0
fix
+ A.x)
Av
-fix)
lim
ix +
Axf-
A.X
-x"
lim
A.v->0
3x^Ax + 3.r(Ax)^ +
(Ax)^
Ax
= lim (3x^ + 3xAx + (Ax)^) = 3x^
At (2, 8), the slope of the tangent \s m = 3(2)- = 12.
The equation of the tangent line is
> - 8 = 12(;t - 2)
V = 12;>: - 16.
(b)
V
\/-
/
(b)
r
/
1
29. (a) fix) = Vx
■' AX-.0 Ax
= lim
Vx + Ax - V^ Vx + Ax + Vx
z^mo Ax Jx + Ax + J~x
(x + Ax) - X
= lim
Aat-^o A,x( V.x + Ax + Vx)
= lim
1
A:c->o Vx: + Ax + Jx iji
At (1, 1), the slope of the tangent line is
1 1
m = — 1= = -.
2j\ 2
The equation of the tangent line is
y-l=\ix-i)
1 1
V = tj: + T.
■22
(b)
56 Chapter 2 Differentiation
31. (a)/(;c) =x + -
f'ix) = lim
f{x+ ^x)-f{x)
Ax
= lim
Ax->0
{x + Ax) + ——r - {x + -
x + Ax \ X
= lim
lim
Aj:->0
Ax
x{x + Ax)(x + Ax) + 4x - x\x + Ax) - 4(.t + A^-)
x{Ax)(x + Ax)
x^ + 2x^(Ax) + x(Ax)^ - x^ - x\Ax) - A(Ax)
x{Ax)(x + Ax)
x^(Ax) + ^(Ajc)^ - 4(M
Ajuo x{Ax){x + Ax)
,. x^+x{Ax)-A
Imi — ; — T —
Ajv^O x(x + Ax)
= 1
4
X X
At (4, 5), the slope of the tangent line is
,43 : :.
16 4
The equation of the tangent line is
(b)
y
1 J/'
X^\
y - 5 = -(x - 4)
y = -^x + 2.
33. From Exercise 27 we know that/'U) = 3x^. Since the
slope of the given line is 3, we have
3^2 = 3
x = ±\.
Therefore, at the points (1,1) and (- 1, - 1) the tangent
lines are parallel to 3x - > + 1 =0. These lines have
equations
and y + 1 = 3(.t + 1)
y-l=3{x- 1)
y = 3x - 2
y = 3x + 2.
35. Using the limit definition of derivative,
■1
fix) =
2xVx'
Since the slope of the given line is —5, we have
1_^ _1
2x^/x 2
x= 1.
Therefore, at the point (1, 1) the tangent line is parallel to
x + 2y - 6 = 0. The equation of this line is
1,
y -
1 = --{x - 1
y -
1 1
l = -2- + 2
1 ^3
37. ^(5) = 2 because the tangent line passes through (5, 2).
gXi)
2-0 _ 2
5 - 9 ~ -4
39. f{x) = X =>f'ix) = 1 matches (b)
Section 2. 1 The Derivative and the Tangent Line Problem 57
41. f(x) = Jx^> f'{x) matches (a)
(decreasing slope as jr — > oo)
43.
Answers will vary.
Sample answer: y = —x
45. (a) If /'(c) = 3 and /is odd. then/'(-c) =/'(c) = 3
(b) If /'(c) = 3 and/ is even, then/'(-c) = -/(c) = -3
47. Let (.rg, y^) be a point of tangency on the graph of/. By the limit definition for the derivative, /'(j;) = 4 - Ir. The slope of the
line through (2, 5) and (xq, Vq) equals the derivative of/ at x^.
5 - v„
= 4 - 2Xn
1- Xq
5 - Vo = (2 - x^){A - 2xo)
5 — {4xq — Xff) = 8 — 8xg + Zkq-
0 = Xq- - 4xo + 3
0 = (.Vo - l){xo - 3) =
J^o = 1.3
Therefore, the points of tangency are (1, 3) and (3, 3), and the corresponding slopes are 2 and -2. The equations of the tangent
lines are
y-5 = 2(.r-2) y - 5 = -2(.i: - 2)
y = 2.V + 1 y = -2x + 9
49. (a) g'iO) = -3
(b) ^'(3) = 0 ■ • ■
(c) Because g'(l) = -3, g is decreasing (falling) at x = 1.
(d) Because g'(-4) = 3, g is increasing (rising) at x = —4.
(e) Because g'{4) and g'(6) are both positive, g(6) is greater than g{4), and g(6) - g(4) > 0.
(f) No, it is not possible. All you can say is that g is decreasing (falling) at x = 2.
51. fix) = ir^
By the limit definition of the derivative we have/'(.r)
X
— 2
-1.5
-1
-0.5
0
0.5
1
1.5
2
fix)
-2
27
32
1
4
1
32
0
1
32
1
4
27
32
2
fix)
3
27
16
3
4
3
16
0
3
16
3
4
27
16
3
2
58 Chapter 2 Differentiation
53. gix)
fix + Om)-f{x)
0.01
= {2(x + 0.01) - {x + 0.01)2 _ 2x + a;2) + 100
3
\
^^
' /
\\
The graph of g(x) is approximately the graph off'{x).
55. /(2) = 2(4 - 2) = 4, /(2.1) = 2.1(4 - 2.1) = 3.99
T.qq — A
f'{2) - Yl^ = ~°-^ [Exact: /'(2) = 0]
57.f{x)=^andf'(x)=:^.
As X —> oo, / is nearly horizontal and thus /' ~ 0.
59. fix) = 4 - (;c - 3)2
/(2 + Ax)-/(2)
■Sa^ (x) =
iu
{x-2)+f{2)
4 - (2 + Ax - 3)2 - 3, ^, ., 1 - (Ax - 1)2 ^^ , , , ^., ^. ,
^ (jc - 2) + 3 = 4 H-t - 2) + 3 = i-^x + 2){x - 2) + 3
Ajc
Ax
(a) Ax = 1: 5^ = (x - 2) + 3 = X + 1
Ac = 0.5: S
^- (-j(x-2) + 3 =-x
19
Ax = 0.1:5,,= ^-j(x-2) + 3=-x--
(b) As Ajc->0, the line approaches the tangent line to/at (2, 3).
>
/^
^/ \
in. -'
61. /(x) =x2 - 1,C = 2
f'Oi 1- /W-/(2) ,. (x2 - 1) - 3 ,. (x - 2)(x + 2)
/ (2) = lim = lim — ■ — — = lim —
x->2 X — 2 x->2 X — 2 x->2 X — 2
= lim (x + 2) = 4
x->2
63. /(x) =x3 + 2x2+ l,c= -2
/(x) -/(-2) ,. x2(x + 2) ,. (x^ + 2x2 + 1) - 1
/'(-2) = lim ^^-^^^^ — ^-r = lim "" ^"" ' ' = lim
x->-2 X + 2 .r->-2 X + 2 x-4-2
x + 2
= lim x2 = 4
x^-2
65. g(x) = Vjlf, c = 0
pfr) - p(0) VlxT
^'(0) = lim ^^ fr^ = lim —!-!■. Does not exist.
J-^O X — 0 x->0 X
As X ^ 0 , = — p — > - oo
^ Vx
yui 1
Asx^ 0-^,—'-^ = ^^ oo
X Vx
67. /(x) = (x - 6)2/3, c = 6
j:— *6 X ~ O
,. (X - 6)2/3 _ 0
= am ;:
x-^6 X — 6
= lim-
1
""e (x - 6)'/3
Does not exist.
Section 2. 1 The Derivative and the Tangent Line Problem 59
69. h{x) = U + 5),c = -5
V(-5)= lim^-W^^
-t->~5 X - ( — 5)
,. |;c + 5| - 0
= lim ■■ ■— —
x->-5 X + 5
= lim
k + 5|
-5 X + 5
Does not exist.
71. fix) is differentiable everywhere except at ;<:
(Sharp turn in the graph.)
-3.
73. fix) is differentiable everywhere except at j: = - 1 .
(Discontinuity)
75. fix) is differentiable everywhere except at j: = 3.
(Sharp turn in the graph)
77. fix) is differentiable on the interval (1, oo).
(At j: = 1 the tangent line is vertical)
79. fix) is differentiable everywhere except at .r = 0.
(Discontinuity)
81./(;c)= |;t- 1|
The derivative from the left is
lim^W^=lim^^
The derivative from the right is
lim ^«^ = lim ^^
Jr-»1* X — I Jr-»1* X — 1
= 1.
The one-sided limits are not equal. Therefore, /is not
differentiable atx = 1 .
r<^ + 1, r < 2
85. Note that/is continuous at x = 2. fix) = \ ^
[4x - 3, -v > 2
83. fix)
ix - 1)3, X < 1
[x - 1)-, X > 1
The derivative from the left is
lim
/U)-/(i)
X - I
lim
ix - 1)
X - 1
lim ix - 1)= = 0.
The derivative from the right is
j:-»1* .t - 1 j:-»1" X - 1
= lim ix - 1) = 0.
Jr->1*
These one-sided limits are equal. Therefore. /is
differentiable at .t = 1. (/'(I) = 0)
The derivative from the left is lim ■'^-^ — {^ = lim — = lim (:t 4- 2) = 4.
Jr->2- X — 2 jr-»;- A" — 2 jr->2-
The derivative from the right is lim — '■ — = lim — '■ = lim 4 = 4.
X-.2* X - 2 x-*2* X - 2 .r->2*
The one-sided limits are equal. Therefore, /is differentiable at v = 2. (/'(2) = 4)
87. (a) The distance from (3, 1) to the line /?ir — >> -I- 4 = 0 is
|Ati + By, + C\
J A- + B'
|/7i(3) - 1(1) -F 4| |3m4-3|
(b)
JirF+l.
Jm- + 1
The function d is not differentiable at m = — 1. This corresponds to the line
y = —X + 4, which passes through the point (3, 1).
60 Chapter 2 Differentiation
89. False. The slope is lim
A;e->0
/(2 + Ax)-/(2)
^x
91. False. If the derivative from the left of a point does not equal the derivative from the right of a point, then the derivative does
not exist at that point. For example, if/(x) = \x\, then the derivative from the left at j: = 0 is - 1 and the derivative from the
right at a: = 0 is 1 . At j£ = 0, the derivative does not exist.
93. f(x) =
xsin(l/jc), X i^ 0
0,
x = 0
Using the Squeeze Theorem, we have -\x\ < x sin(l/x) < |a:|, a: =?^ 0. Thus, lim x sia(\/x) = 0 = /(O) and/is continuous at
X = Q. Using the alternative form of the derivative we have
,. fix) -/(O) ,. A:sin(lA) - 0 ,. , .
lim ''-^^ — 77^ = Imi ^"-^ = lim I sm
x-iO X — \J x^O X — \J J:->0
"i)-
Since this limit does not exist (it oscillates between - 1 and 1), the function is not differentiable at a: = 0.
g{x)-
X^ SV!\i\/x), X T^ 0
0,
0
Using the Squeeze Theorem again v.'e have -x'- < x'- sia(l/x) < x^,x i= 0. Thus, lim :? sin(l/j:) = 0 = /(O) and/is continu-
ous at X = 0. Using the alternative form of the derivative again we have
,. /W-/(0) ,. x2sin(lA)-0 ,. . 1 „
lim -''^-^ — r^ = lim —'—T = hm x sin - = 0.
jr^O X — 0 j:->0 X — 0 x-^0 X
Therefore, g is differentiable at x = 0, ^ '(0) = 0.
Section 2.2 Basic Differentiation Rules and Rates of Change
1. (a)
y = x'/2
yd) =5
3. y = 8
y' = Q
(b) y = x3/2
„l/2
y'W = I
5. >' = x6
y'= 6x?
(c)
7. y
y = x^
(d) y = ^
y' = 2x
y'=3x^
y'il) = 2
y\\) = 3
x' ""
9. y = ^ = xi/5
= -7x-8 =
-7
y 5 5x^/5
11. fix) = X + 1
fix) = 1
13. fit) = -2;2 + 3r - 6 IS. g(x) = x^ + 4x3
/'(;)= -4, + 3 g'W = 2x + 12x2
17. sit) = t^-lt + A
s'ii) = 'hi^ -1
77
19. y = ^ sin 6 - cos e
21. y = x^ — X cos X
23. V = 3 sin X
x
y' = —cos e + sin 0
y' = 2x + — sinx
x2
— 3 cos X
Section 2.2 Basic Differentiation Rules and Rates of Change 61
Function Rewrite
25. V
27. y =
29. y
3
(2x)3
>" = ?
>' = 7-^
>> = x
-1/2
Derivative
y'= -5;c-3
y' = -5^-^"''
= --r-3/2
Simplify
y =
^
2x^/2
31. fix) = ^ = 3x-2, (1, 3)
fix) = -6x-3 = -J-
/'(l) = -6
33. fix) = -| + |x3, (o, -^
21
Z'U) = yx2
/'(O) = 0
35. V = (2x + 1)2, (0, 1)
y'= 8;c + 4
j'(0)=4
37. /(e) = 4 sin 0 - e, (0, 0)
fie) = 4 COS e - 1
/'(O) = 4(1) -1=3
39.
fix) = x'- + 5 - 3^-2
/V) = 2x + 6x-3 = 2i +
41. ^(r) = ^2 - -^ = f2 - 4r3
g'(f) = 2f + 12r-* = 2r +
12
43. fix) = ^^ ^^^^^ = ;c - 3 + 4.t-2
/'W = l-3 ""
X3 ^3
47. /(x) = v^ - 6 ^ = xi/2 - 6x'/3
1 1 2
51. fix) = 6Vx + 5 cos x = 6.x:'^2 + 5 cosjc '
/'(x) = 3x ''2 — 5 sin JT = — = - 5 sin jc
45. y = x(x2 + 1) = x3 + X
y'= 3x2 + 1
49. his) = i-*/^ - ^2/3
4 9 4 7
t Y^l = -5-4/5 _ £c-i/3 = —Z f_
53. (a) >- = X* - 3x2 + 2
y' = 4x3 - ^x
At (1.0): y' = 4(1)3 _ 5(1) = _2.
Tangent line: y - 0 = -2(x - 1)
2x + y - 2 = 0
(b)
v;
V
\/^
(1.0?^
55. (a) fix) = ^ = ^1^"'^"
-3 —3
f'(x) = ^x"''/" = — —
At (1, 2), /'(I) = -^
Tangent line;
V - 2 = -Hx - 1)
3 . 7
3x + 2v - 7 = 0
y = --X + -
(b)
\
\^_
\
62 Chapter 2 Differentiation
57. y = X* - Sx^ + 2
>>' = 4j:^ - 16jr
= 4x{x^ - 4)
= 4x{x - 2){x + 2)
y' = 0 => j: = 0, ±2
Horizontal tangents: (0, 2), (2, - 14), (-2, - 14)
59. y
x2
2x ' = — ^ cannot equal zero.
Therefore, there are no horizontal tangents.
61. y = X + sin X, 0 < X < Itt
y ' = 1 + cos X = 0
cos JC=— 1 => X = TT
At X = TT, y = TT.
Horizontal tangent: (n, n)
63. AT^ — fcc = 4x - 9 Equate functions
2x - k = 4 Equate derivatives
Hence, k = 2x — 4 and
x^ - {2x - 4)x = 4x - 9-
For jc = 3, <: = 2 and for x
-x^ = -9=:>j: = ±3.
-3, /t= -10.
A; 3
65. - = — -X + 3 Equate functions
X 4
k^
X-
Equate derivatives
3 2
3 4^^
Hence, k = —x^ and —
4 a:
-3^1 3
4 4
3.3.,
■ — X + 3 =» -X = 3 :
. X = 2 =» A: = 3.
67. (a) The slope appears to be steepest between A and B.
(b) The average rate of change between A and B is
greater than the instantaneous rate of change at B.
(c)
69. g(x)=/(x) + 6^g'W=/'(x)
71,
ff/is linear then its derivative is a constant function.
fix) = ax + b
fix) = a
Section 2.2 Basic Differentiation Rules and Rates of Change 63
73. Let (x^, y^ and (xj, y^^ be the points of tangency ony = x^ and y = —x~ + 6x - 5, respectively. The derivatives of
these functions are
y' = 2x =^ m = 2x^ and y'= — 2x + 6 => m= — Ixj + 6.
m = 2x, = — Ivj + 6
JCl = -^2 + 3
Since y, = .r,^ and jj = -x^^ + 60:3 ~ 5,
V2 - yi (-x,^ + 6x2 - 5) - (-x,^)
ra = = = — 2x, + 6.
{-x.} + 6X-, - 5) - (-x, + 3)2
= -2x, + 6
X2 — {~X2 + 3)
{-x^ + 6x2 - 5) - Uj- - 6x2 + 9) = (-2x2 + 6)(2x2 - 3)
-Ixj' + 12t2 - 14 = -4x2- + 18x2 - 18
2x2' - 6x2 + 4 = 0
2(x2 - 2)(x2 - 1) = 0
1 or 2
1 => y2 = 0, X, = 2 and Vj = 4
Thus, the tangent line through (1,0) and (2, 4) is
y-0 =
(x - 1) => y = 4x - 4.
\2 - \)
Xj = 2 => y, = 3, X, = 1 and Ji = 1
Thus, the tangent line through (2, 3) and (1, 1) is
(i - r
V - 1 =
2 - 1
(x - 1) => y = 2x - 1.
75. /(x) = J~x, (-4,0)
'■"'-r"'-in
1
0 - V
ij'x -4-x
4 + X = 2Vxy
4 + X = 2VxVx
4 + X = 2x
X = 4, y = 2
The point (4, 2) is on the graph of/.
0-2
Tangent line: y - 2
-4-4
4y - 8 = X - 4
0 = X - 4v + 4
(.t - 4)
77. /'(I) = - 1
64 Chapter 2 Differentiation
79. (a) One possible secant is between (3.9, 7.7019) and (4, 8):
8 - 7.7019,
y-i
<^-4)
4 - 3.9
y - 8 = 2.981U - 4)
y = 5W = 2.981;c- 3.924
(b) fix) = \x^l^ ^ /'(4) = f(2) = 3
r(.r) = 3(;c - 4) + 8 = 3jc - 4
5(x) is an approximation of the tangent line Tt^:).
(c) As you move further away from (4, 8), the accuracy of the approximation T gets worse.
(d)
Ax
-3
-2
-1
-0.5
-0.1
0
0.1
0.5
1
2
3
/(4 + Ax)
1
2.828
5.196
6.548
7.702
8
8.302
9.546
11.180
14.697
18.520
7t4 + Ax)
-1
2
5
6.5
7.7
8
8.3
9.5
11
14
17
81. False. Let/(x) = x- and g{x) = x^ + 4. Then
/'(x) = g'W = 2x,but/(x)^g(x).
83. False. If y = ir^, then dy/dx = 0. (tt^ is a constant.)
85. True. If g(x) = 3/(.x), then g'(x) = 3/'(x).
87. f(t) = 2t + 7, [1,2]
fit) = 2
Instantaneous rate of change is the constant 2.
Average rate of change:
/(2)-/(l) [2(2) + 7] - [2(1) + 7]
= 2
2-1 1
(These are the same because/is a line of slope 2.)
89. f(x) = --. [1, 2]
fix)
x'
Instantaneous rate of change:
(1,-1) ^/'(1) = 1
2.-l)^rm-\
Average rate of change:
/(2)-/(l) (-1/2) -(-1) ^ j_
2-1 2-1 2
Section 2.2 Basic Differentiation Rules and Rates of Change 65
91. (a) s{t) = - 16r= + 1362
v(r) = -32r
sir) - s{\)
^"> 2-1 " ^^^^ - 1346 = -48 ft/sec
(c) v(t) = sXt)= -32/
Whenf = 1: v(l) = -32 ft/sec.
When t = 2: v(2) = -64 ft/sec.
(d) - 16/2 + 1362 = 0
1362
f2 =
16
/ = ^^» 9.226 sec
(e)v(^) = -32(^)
= -871362 « -295.242 ft/ sec
93. s(t) = -A.9fi + Vot + So
= -4.9f2 + 120r
v(r) = -9.8; + 120
v(5) = -9.8(5) + 120 = 71 m/sec
v(10) = -9.8(10) + 120 = 22 m/sec
95.
■S. 50--
.S '»
-t— I-
2 4 6 8 10
Time (in minutes)
(The velocity has been converted to miles per hour)
99. (a) Using a graphing utility, you obtain
« = 0.167V - 0.02.
(c) T= R + B = 0.00586V- + 0.1431v + 0.44
(e) -- = 0.01172V + 0,1431
flv
Forv = 40, r'(40) « 0.612.
Forv = 80, r'(80) « 1.081.
Forv = 100, r'(lOO) =« 1.315.
97. V = 40 mph = f mi/min
(f mi/min)(6 min) = 4 mi
V = 0 mph = 0 mi/min
(0 mi/min)(2 min) = 0 mi
V = 60 mph = 1 mi/min
(1 mi/min)(2 min) = 2 mi
(b) Using a graphing utility, you obtain
B = 0.00586v2 - 0.0239V + 0.46.
(d) 60
Time (in minules)
(f) For increasing speeds, the total stopping distance
increases.
101. A
, dA
' ds
When .r = 4 m,
dA
ds
= 8 square meters per meter change in i.
103.
^ 1,008,000 ,,^
C = :; + 6.3(2
dC_
dQ
Q
1,008.000
+ 6.3
C(351) - C(350) = 5083.095 - 5085 = -$1.91
d€
When Q = 350,
dQ
$1.93.
105. (a) /'(1 .47) is the rate of change of the amount of gasoline sold when the price is $1.47 per gallon,
(b) /'(1. 47) is usually negative. As prices go up, sales go down.
66 Chapter 2 Differentiation
107. y = ax- + bx + c
Since the parabola passes through (0, 1) and (1, 0), we have
(0, 1): 1 = a(0)2 + b{Q) + c ^ c = 1
(1, 0): 0 = a{\Y + b(\) + 1 =* fc = -a - 1.
Thus, y = ax^ + {—a — \)x + \. From the tangent line y = x - 1, we know that the derivative is 1 at the point (1, 0).
y' = lax + {-a — 1)
1 = 2a(l) + (-«-!)
1 =a - 1
a = 2
6= -a - 1 = -3
Therefore, y = 2x^ - l,x + \.
109. y = ]^ -9x
y' = 3x2 - 9 ■ ,. .
Tangent lines through (1, —9):
. y + 9 = (3x2 _ 9)(^ _ 1)
• (x3 - 9x) + 9 = 3x3 - 3x2 - 9x + 9
0 = 2x3 - 3x2 = ^2(2;c - 3)
X = 0 or X = 2
The points of tangency are (0, 0) and (f, -y). At (0, 0) the slope is ^'(0) = -9. At (f, -y) the slope is y '(2) = "I •
Tangent lines:
y-Q= -9(;c - 0) and y + f = "K^ "2)
n 9 27
y = -9x y = -4X - -^ ■ ■
9x + J = 0 9x + 4y + 27 = 0
111. /(x) = 2 . J, ^ o
[x2 + i), X > 2
/must be continuous at x = 2 to be differentiable at x = 2.
lim fix) = lim ox^ = 8a 1 8a = 4 + fo
lim /(x) = lim (x2 + /,) = 4 + Z7 J ^"^ ~ ■* " ^
x->2* x-»2*
fix)
3ax2, X < 2
2x, X > 2
For/to be diiferentiable at x = 2, the left derivative must equal the right derivative.
3a(2)2 = 2(2)
12a = 4
a = 3
i = 8a - 4 = -I
Section 2.3 The Product and Quotient Rules and Higher-Order Derivatives 67
113. Let/W = cosx
f(x + Ax)-f{x)
f'(x) = lim
Ax->0
= lim
Ax-»0
Ax
cos X COS Ajc — sin j: sin Ajc — cos x
Ajc
,. cosx(coszix- 1) ,. /sinAx
= liin : lim smx — : —
Ax->o Ajc /u->o \ Ax
= 0 - sinx(l) = -sinx
Section 2.3 The Product and Quotient Rules and Higher-Order Derivatives
1. g(x) = (x- + DU^ - It)
g'(x) = {x'- + l)(2.r - 2) + ix^ - 2x)(2x)
= Ix^ - 2^2 + 2r - 2 + 2x3 - 4x2
= 4x^ - 6x- + lx - 2
S. f(x) = x^cosx
f'(x) = x3(— sinx) + cosx(3x-)
= 3x- cosx — x^ sinx
3. hit) = yiifi + 4) = fi/3(f2 + 4)
h'(t) = f'/3(2f) + (f2+4)l -2/3
= 2f4/3 +
t- + 4
3f2/3
7f2 + 4
3/2/3
7. /(x)
X2 + 1
fix) =
(x2 + 1)(1) - x(1t) 1 -x2
(x2 + 1)2 (x2 + 1)2
9. h(x)
rx
rl/3
X3 + 1 .x3 + 1
h\x) =
(X3 + l)jX-2/3 - X'''3(3x2)
U^Tl)^
(x3 + 1) - X(9x2)
3x2/3(x3 + 1)2
1 - 8x3
3x2/3(x3 + 1)2
11. g(x
gXx) =
x2(cosx) - sin.t(2t) _ x cos x - 2 sin x
(x2)2 - x3
13. fix) = (x3 - 3x)(2t2 + 3x + 5)
fix) = (x3 - 3x)(4x + 3) + (2x2 + 3;c + 5)(3;c2 _ 3)
= lOx^ + 12x3 - 3^2 _ 18^ _ 15
/'(0) = -15
15. fix) =
/'W =
/'(I) =
x2 - 4
X - 3
ix - 3)(2x) - (x2 -
-4)(1)
2r2 - 6.r - x^
U - 3)2
* 4
U - 3)2
.r2 - dx + 4
U - 3)2
1-6+4 1
(1 - 3)2 4
17. fix) = X cos X
fix) = (x)(-sinx) + (cosx)(l) = cosx - xsinx
Hf)=#-f(#)=f'-'»
68 Chapter 2 Differentiation
Function Rewrite
Of
nvafive
19. v^^^;^ .^.^^f.
y'
2 2
= 3^^3
21.. = 3^3 . = 1-^
y'
= -7;c-4
23. y = — y = 4v^, ;c> 0
y'
= 2;c-'/2
-/« = ^^^
.. . U^ - l){-2 - 2;c) - (3 - 2;c - x^)(2x)
/W- (^2_l)2
2^2 - 4;c + 2 2(x - 1)^
"U+D-^-'^i
/ 4 \ 4x
27./W=.(l-^^3)=.-^^3
29.
(x + 3)4-4x(l) U2 + 6;c + 9)-
-^^""^ U + 3)2 " (;c + 3)2
12
^2 + 6;c - 3
Simplih
, 2;c + 2
2a: + 5
2;c'/2 + 5;c-i
fix) =X-'/2-|;C-3/2 = X-3/2
^~i
U + 3)2
2;c - 5 ^ 2x- 5
2^Vx " 2x3/2
31. ftW = (53 - 2)2 = 56 - 4^ + 4
/i'(i) = 6^ - 12*2 = 6^2(^3 - 2)
33. fix)
2x- \ 2x- \
X — 3 xix — 3) ^2 — 3x
, , ^ U2 - 3a:)2 - (2x - l)(2x - 3) ^ 2^2 - 6x - 4^2 + 8a - 3
^ ^''' U2 - 3a)2 ^2 - 3x)2
-2^2 + 2x - 3
(x2 - 3x)2
2^2 - 2x + 3
x2(x-3)2
35. /W = (3jt3 + 4x)ix - 5)(x + 1)
fix) = (9x2 + 4)(^ _ 5)(^ + 1) + (3x3 + 4x)i\)ix + 1) + (3x3 + 4^)(^ _ 5)(i)
= (9x2 + 4)(^2 _ 4^ _ 5) + 3^4 + 3^3 + 4^2 + 4;c + 3;c4 _ i5_^ + 4j^2 _ 20x
= 9x* - 36x3 _ 4i;f2 _ 15^ - 20 + 6x^ - 12x3 + 8x2 - j^^
= ISx'* - 48x3 - 33x2 _ 2,2x - 20
37. /(x) =
/'(^) =
X2 + C2
X2-C2
(x2 - c2)(2x) - (x2 + c2)(2x)
(x2 - c2)2
39. fix) = fsmt
fit) = P cost + 2t sin t
= titcost + 2 sin t)
—4x^2
(x2 - c2)2
Section 2.3 The Product and Quotient Rules and Higher-Order Derivatives 69
41. fit) = ^ 43. fix) = -X + tan ;t
„/ V _ -tsinf - cosf _ f sin t + cos t
fix) = - 1 + sec^j: = laiv^x
45. g(f) = 4/^ + 8 sec f = f'/" + 8 sec f 47. y = ^''\ ^'"''^ = |(sec.t - tanx)
-^ 2cos.r 2
g'W =7^3'"' + 8secftanf = — J7T + Ssecrtanf ,3, , , 3 ,
^ ^r"' ;y = -(sec j: tan .t - sec- x) = Tsec .r(tan x - sec x)
3
= -(sec X tanx — tan^x - 1)
49. y — — CSC X — sin X 51. fix) = x^ tan x
y ' = CSC X cot X — cos x fix) = x- sec- x -H 2x tan x
COS X = x(x sec^ X -H 2 tan x)
= —r^ cos X
sin-x
= cosx(csc-x — 1)
= cos X cot'^ X
53. y = 2t sin X + x^ cos x 55. g(;^) = |1±_1 W _ 5)
>> ' = 2x cos X + 2 sin X -I- x-(— sin x) -t- 2x cos x
= 4x cos X -^ 2 sin x - x^ sin x S 'W = ^, , .^2 tfo™ of answer may vary)
(x + 2)2
57. g(e) =
1 — sin
,, . 1 - sin 9 -I- e cos e
ff (0) = r^ — Tz:; (form of answer may vary)
(sm 9-1)-
1 + CSC X "
59. y =
1 — CSC X
- , _ (1 - cscx)(-cscxcotx) - (1 -I- CSC x)(csc X cot x) _ -2 CSC X cot X
-*' ~ (1 - cscx)= ~ (1 - cscx)-
M ^ -2(2)(^) ^ _^
y[6) (1-2)2 4V3
61.
ftW =
sec t
f
h'it) =
f(sec r tan f) -
(sec r)(l)
•
t-
=
sec r(f tan r —
;2
i)
/iV) =
sec Triirtan tt
—
il
1
70 Chapter 2 Differentiation
63. (a) f(x) = {x^ -3x+ l){x + 2), (1, -3)
fix) = (;c3 - 3.t + 1)(1) + {x + 2)(3;c2 - 3)
= 4;c3 + 6;c2 - 6x - 5
/'(I) = -1 = slope at (1,-3).
Tangent line: y + 3 = —l(x- 1) => y = —x — 2
(b)
65. (a) fix) =
fix) =
/'(2) =
(2, 2)
;c - 1
(x - 1)(1) - xil) -1
ix - 1)2 ix - 1)2
1
= - 1 = slope at (2, 2).
(2 - 1)2
Tangent line: >> - 2 = - lU — 2) => 3? = -x + 4
(b)
-i^
\i \
67. (a) /W = tanx, (^ j, 1
/'(jc) = sec2x
/'(f) = 2 = slopeat Jl
Tangent line:
y-i=2[x
(b)
jm
J(
if
Y
y-i = 2x--
4j:-2y-Tr+2 = 0
69
/(-t) =
JC -
1
fix) =
ix-
- 1)(2x) - ;.
^(1)
ix - 1)2
^2
- 2x x(jc
-2)
ix - 1)2 (x - 1)2
fix) = 0 when x = 0 or x = 2.
Horizontal tangents are at (0, 0) and (2, 4).
71. fix) =
g'ix) =
(x + 2)3 - 3x(l) _ 6
ix + 2)2 (x + 2)2
(x + 2)5 - (5x + 4)(1) 6
ix + 2)2
ix + 2)2
, , 5x + 4 3x 2x + 4 ,, , ^
/ and g differ by a constant.
73. /(x) =x"sinx
fix) = x" cos X + nx" " ' sin X
= x"" ' (x cos X + n sin x)
When n = 1
When n = 2
When n = 3
When n = 4
/'(x) = X cos X + sin x.
/'(x) = x{x cos X + 2 sin x).
fix) = x2(x cos X + 3 sin x).
fix) = x^ix cos X + 4 sin x).
75. Area = Ait) = (2f + l)Vf = 2/3/2 + ^1/2
A'W = 2(|f'/2)+|r'/2
= 3?i''2 + 1,-1/2
6t + 1
2Vf
cm2/sec
For general n,f'ix) = x" ' (x cos x + n sin x).
Section 2.3 The Product and Quotient Rules and Higher-Order Derivatives 71
77. C= 100(202 +
X
""^ ioo(-^ +
■2 X + 30
30
1 < X
dx
(a) When ;c = 10
:^ {x + 30)2
dC
dx
dC
$38.13.
(b) Whenx = 15: -— = -$10.37.
dx
/"ft"'
(c) When a: = 20: -— = -$3.80.
dx
As the order size increases, the cost per item
decreases.
79. Pit) = 500
P'it) = 500
= 500
= 2000
11 '*' "
L 50 + f2j
(50 + r2)(4) -
- (4r)(2r)
(50 + Pjz
200 - 4t-
(50 + t^y_
" 50 = r
.(50 + a^J
P'(2) = 31.55 bacteria per hour
81. (a) sec x =
d. -. d
(cos;c)(0) - (1)(— sinx) _ sin.r _ 1 sin;c
(cos xy
cos X cos x cos X cos X
= sec X tan x
(b) CSC j:
sinx
-[csc^] =
d
dx
1
, sin X
(c) cot X =
cosx
sin.v
-[cot..] =
d
dx
"cos-t
.sin;c
83. fix) = 4x^/2
fix) = 6;ci/2
/'W = 3;c-/2
■N
3
fx
(sin.v)(0) - (1)(cosj:) cosjt
1 cos j:
(sin jc)-
sin X sm X sm x sin x
- CSC X cot x
cos-t] sin .t( — sin.r) — (cos. r)(cosj:) sin^jc + cos^x 1
(sin .r)-
sm''j:
85. fix) =
• ■ /W =
sin-.v
X -
- 1
ix
- 1)(1) -
.t(l)
-1
(-r - D-
ix
-D-
2
ix - D'
87. fix) = 3 sin x
fix) = 3 cos X
f'\x) = - 3 sin X
89. fix) = x^
fix) = 2r
= -) /:
91. /"(x) = 2v'x
/'•'>(.v)=^(2).r-'- = -Jp
93.
/(2) = 0
One such function is/(.v) = (x — 2)-.
95. fix) = 2gix) + hix)
fix) = 2g'ix)+h'ix)
fi2) = 2g'i2) + h'il)
= 2(-2) + 4
= 0
97. fix) =
fix) =
/'(2) =
hix)
hix)g'ix) - gix)h'ix)
[hix)y
h(2)g'i2) - g(2)h\2)
[h(2)Y
(-l)(-2)-(3)(4)
(-1)-
= -10
72 Chapter 2 Differentiation
99.
It appears that/ is cubic; so/' would be quadratic and
/"would be linear.
103. v(r)
, a(t)
lOOr
2t+ 15
(It + 15)(100) - (100?)(2)
{2t + 15)2
1500
(2r + 15P
(a) a(5) =
101. v(t) = 36 - r^, 0<r<6
a(r) = -2t
v(3) = 27 m/sec
a{3) = - 6 m/sec
The speed of the object is decreasing.
1500
(b) a(10) =
(c) a(20) =
[2(5) + 15?
1500
[2(10) + 15?
1500
[2(20) + 15]^
= 2.4 ft/sec2
= 1.2ft/sec2
» 0.5 ft/sec2
105. f(x) = g(x)h{x)
(a) fix) = gix}h'{x) + h(x)g'(x)
fix) = g{x)li"{x) + g'(x)h'{x) + h(x)g"ix) + h'{x)g'ix)
= g(x)hXx) + 2g'{x)h Xx) + h{x)g\x)
rix) = g(x)h"'(x) + g'{x)h"(x) + 2g'{x)h'\x) + 2g"ix)h'(x) + h(x)g"'(x) + h'{x)g%x)
= g{x)h '"{x) + 3g 'ix)h'{x) + 3g%x)h Xx) + g "'ix)hix}
f^%x) = g(xW'Kx) + gXx)h"Xx) + 3gXx)h"Xx) + 3g'\x)h'U) + Sg'UMx) + 3g"'{x)hXx)
+ g"Xx)hXx) + g^'KxMx)
= gixW'Kx) + 4gXx)h"Xx) + 6g'{x)h'{x) + 4g"'ix)hXx) + g^'Kx)h{x)
(X>) f w gwn W + j^(^ _ j)(^ _ 2) . . . (2)(1)^ ^'''^ ^''' ^ (2)(l)[(n - 2)(n - 3) ■ ■ • (2)(1)]^ ^' ^'
+ ,.J:[-:^^z'^-::^^'L.,g"'ix)h^-^^(^ + ■ ■ ■
(3)(2)(1)[(« - 3)in - 4) ■ ■ • (2)(1)
n{n- \)(n-2)- ■ ■ (2)(1)
[{n-l)(n-2)- ■ •(2)(l)](i;
g^"-'Kx)hXx)+g^"\x)h{x)
= gix)hMix) + !,(„": i),gW^"'-"(^) + 2!(/- 2)!g'^^)^^"""(^) + ■ ■ ■
-g(«-'>(x)A'(;t) + gW(x);i(;c)
(n- 1)!1!'
Note: «! = «(« — 1) ... 3 • 2 • 1 (read "n factorial.")
Section 2.4 The Chain Rule 73
107. f(x) = cos X
f'(x) = -sinjr
f"(x) = -cos a;
(a) Pi(x) = f'(a)ix - a) + fia
1
^1 7r\ 77 1
/ — = COS — = -
-'* 3/ 3 2
r/| ■''■\ • "■ v/3
/ljj = -s.n-=-—
/(- =-cos-=--
^f. - ^] .
2 V'
= -4l"-3J -T"l"-3J + 2
(c) P, 's a better approximation.
(b)
/■jX
<.
>
(d) The accuracy worsens as you move farther away
fromx = a = (ir/S).
109. False. If y = f{x)gix), then
dy
dx
- f(x)g'{x) + g{x)f'{x).
111. True
h'(c) = f(c)g'(c) + g(c)f\c)
= /(c)(0) + g(c)(0)
= 0
113. True
115. fix) = x\x\ =
X-, iix>Q
-x^, \ix < 0
fix)
fix) =
2x, ifjc > 0
-2x, \ix < 0
2, if a: > 0
-2, if.)c < 0
/"(O) does not exist since the left and right derivatives
are not equal.
Section 2.4 The Chain Rule
y=Mx))
1. y = {6x - 5)"
" = g(x)
u = bx — 5
■ y=f{u)
y = W
3. y = Ti^^nr
u = X- — 1
= ^
5. V = csc'j:
M = cscx
7. .V = (2x - 7)'
y'=3(2j<:- 7)-(2) = 6(2x- 7)^
9. ^(.v) = 3(4 - 9.V)-'
g'(x) = 12(4 - 9Af(-9) = - 108(4 - 9x)^
11. fix) = (9 - ;c=)=/'
13. /(f) = (1 - f)"-
/'(f)=^l-r)-'-(-l)
2vl -/
74 Chapter 2 Differentiation
15. y = (9;(2 + 4)1/3
17. y = 2(4 - j;2)i/'»
^'=2^ (4-;c2)-3'V2;c)
4^(4 - ;c2
19. >> = (;c- 2)-'
y'=-l(2-;c)-2(l) =
1
(^ - 2?
21. /(t) = (r - 3)-2
fit) = -2(r - 3)-3 =
a - 3)3
23. y = (x + 2)-'/2
^ = _i(^ + 2)-3/2 = \
25. f{x) = x\x - If
fix) = x\A{x - 2)3(1)] + (;c - 2)^(2;t)
= 2x{x - 2)3[2x + (x- 2)]
= 2x{x - 2)3(3x - 2)
27. y = xj\ - x^ = x{\ - x^y^
y =x
\(\ - x^)-'/^(-Xx)
+ (1 - x2)'/2(l)
= -xW - x^Y"^ + (1 - x2)'/2
= (1 - x')-''\-x^ + (1 - ;c2)]
1 -2x2
yn^
/ < ,' x + 5
31. g{x)= —
gW = 21
;c2 + 2,
;c + 5 \/(x2 + 2) - (x + 5)(2;i;)
35. y
;c2 + 2/\ (a;2 + 2)2
_ l{x + 5)(2 - 10;c - x^)
(x2 + 2)3
J~x + 1
;c2+ 1
1 - 3;c2 - 4;(3/2
275(;c2 + 1)2
The zero of y ' corresponds to the point on the graph of
y where the tangent line is horizontal.
29. y =
= X(x2 + l)-l/2
y' = /-^(x2 + l)-3''2(2x)l + (x2 + l)-'/2(i)
= -;c2(x2 + I)-3/2 + (^2 + l)-l/2
= (;c2 + l)-3/2[-;c2 + {x" + 1)]
1
(;c2 + 1)3/2
33. /(v) =
/'(v) = 3
1 - 2v
1 + V
1 - 2v\V(l + v)(-2) - (1 - 2v)
1 + V / V (1 + v)2
9(1 - 2v)2
37. «(f) =
(1 + v)*
3f2
Jfi + 2t- \
^ Itjfi + 3f - 2)
^ ^ ^ (r2 + 2f - 1)3/2
The zeros of g ' correspond to the points on the graph of
g where the tangent lines are horizontal.
Section 2.4 The Chain Rule 75
39. y
= V^
,^ J(x + l)/x
^ 2x{x + 1)
y ' has no zeros.
. . .^^
L.
v
(
41. s(t)
s\t)
-1{1 - t)J\ + t
JVT
The zero of s '{t) corresponds to the point on the graph of
s{t) where the tangent line is horizontal.
s
43. y
cos TTJC + 1
dy _ — vx sin vx — cos ttx - 1
clx~ x^
TTX sin TTX + cos TTX + 1
M
The zeros of y ' correspond to the points on the graph of y where the tangent lines are horizontal.
45. (a) y = sin X
y' = cosjc
y'io) = 1
1 cycle in [0, Itt]
(b) y = sin 2x
y ' = 2 cos 2x
y'iO) = 2
2 cycles in [0, 2Tr]
The slope of sin ax at the origin is a.
47. y = cos 3jc
dy
dx
= - 3 sin 3a:
51. y = sin (irx)- = sin (ir-x^)
y' = cos (tt x-)[27r\x] = 27r-.vcos(iT-j:-)
53. h(x) = sin 2jc cos 2x
h '(x) = sin 2t(- 2 sin Iv) + cos 2j:(2 cos 2jc)
= 2 cos- 2x — 2 sin- 2t
= 2 cos 4.1:.
Alternate solution: h{x) = — sin 4.r
h '(x) = - cos 4j:(4) = 2 cos 4.r
49. g{x) = 3 tan 4x
g'{x) = 12 sec-4jt
sc f, ^ cot.t cosj:
55. f(x) = = ^^;—
Sin X sin- X
fix) =
sin- .rl— sin .r) — cos xjl sin x cos .t)
sin'* .t
— sin^A: — 2 cos-.r — 1 — cos-.t
sin^ .v
sin- .V
76 Chapter 2 Differentiation
57. y = 4 sec^ X
>> ' = 8 sec a: • sec j: tan ^ = 8 sec^ xtanx
61. fix) = 3sec2(7rf- 0
/'W = 6sec(Trr - l)sec(Trr- l)tan(7rr- IKtt)
6Trsin(Trr — 1)
= 6irsec2(irr - 1) tan(irf - 1) =
cos^ (irt - 1)
59. /(fl) = J sin2 20 = |(sin 26)^
fie) = 2(3)(sin 20)(cos 2e)(2)
= sin2ecos2e = 5 sin 40
63. y = ^ + - sin(2;c)2
= -/x + - sin(4;c2)
£ = |;c-'/2 + icos(4x2)(8x)
1
2v^
+ 2x cos(2x)2
65. y = sin(cosx)
— = cos(cosar) • (— sinx)
ax
= — sin j: cos(cos x)
67. s(t) = it^ + 2t+ 8)'/2, (2,4)
J'W =|(f2 + 2f+ 8)-'/2(2f +2)
t+ 1
^'(2) =
Vr^ + 2r +
3
69- /W = ^ = 3(^3 _ 4)-i, / i,_3
/'W = -3(;c3 - 4)-2(3x2:
9;r2
ix' - 4y-
/'(-1) =
25
71. /(r) = i^, (0,-2)
/'(O
r- 1
(t - 1)(3) - (3f + 2)(1)
-5
(f - 1)^
/'(O) = -5
it - 1)'
73. y = 37 - sec3(2;c), (0, 36)
y' = -3 sec2(2x)[2 sec(2;c) tan(2x)]
= -6sec3(2x)tan(2x)
y'iO) = 0
75. (a) fix) = J3x^ - 2, (3, 5)
/'(x) = ^3;c^ - 2)->/2(6x)
3x
V3]t
/'(3) =
Tangent line:
>< - 5 = -(a: - 3) ^ 9;c - 5y - 2 = 0
(b)
\j
/"
/
77. (a) /(x) = sin 2x, (tt, 0)
fix) = 2 cos 2x
/V) = 2
Tangent line:
> = 2(;c - tt) ^> 2x - y - 27r = 0
(b)
Section 2.4 The Chain Rule 77
79. fix) = 2(;c2 - \f
fix) = 6(;c^ - mix)
= llxU" -2x^+1)
= llr^ - 24A-5 + Xlx
fix) = eOjc" - 72jc2 + 12
= 12(5x2 - l)(jc2 - 1)
81. /(x) = sin x2
/'(x) = 2xcosx2
/"(x) = 2x[2x(-smx=)] + 2cosx2
= 2[cos x^ — Ix^ sin x-]
83.
85.
The zeros of/' correspond to the points where the graph
of/has horizontal tangents.
The zeros of/' correspond to the points where the graph
of/ has horizontal tangents.
87. g(x)=/(3x)
g'(x) =/'(3.r)(3) ^ g'(x) = 3/'(3x)
89. (a) fix) = gix)hix)
fix) = g{x)h'ix) + g'ix)hix)
f'iS) = (-3)(-2) + (6)(3) = 24
('^^^w-fS
fix)
/'(5)
Mx)g'(x) - g(x);it^)
(3)(6) - (-3)(-2) 12 4
(3)2 9 3
(b) /(.r) = gihix))
fix) = g'ihix))h'ix)
/'(5) = g'(3)(-2) = -2g'(3)
Need g'(3) to find /'(5).
(d) fix) = [g(x)]^
/'(x) = 3[g(.r)]2g'(.t)
/'(5) = 3(- 3)2(6) = 162
91. (a) /= 132.400(331 - v)"'
/' = (-1)(132,400)(331 - v)--i-\)
132,400
" (331 - v) =
Whenv = 30,/'= 1.461.
(b) /= 132,400(331 + v)"'
/' = (-1)(132,400)(331 + v)-2(l)
-132.400
(331 + v)-
Whenv = 30,/'= -1.016.
93. e = 0.2 cos 8f
The maximum angular displacement is 9 = 0.2 (since
- 1 < cos 8f < 1).
M
dt
0.2[-8sin8t] = -1.6 sin 8r
When t = 3. dd/dt = - 1.6 sin 24 = 1.4489 radians per
second.
95. S = CiR- - r-)
f=c(2Rf-2r'-f
dt \ dt dt
Since r is constant, we have dr/dt = 0 and
^ = (1.76 X 105)(2)(1.2 X 10--)(10-5)
dt
= 4.224 X 10"- = 0.04224.
78
Chapter 2 Differentiation
97. (a) x= - 1.6372f3 + 19.3120f2 - 0.5082r - 0.6161
(b) C= 60x+ 1350
= 60(-1.6372r3 + 19.3120r2 - 0.5082« - 0.6161) + 1350
dC
dt
= 60(- 4.9 116(2 + 38.624r - 0.5082)
= -294.696r2 + 2317.44f - 30.492
The function — is quadratic, not linear. The cost function levels off at the end of the day, perhaps due to fatigue.
99. f{x) = sin ^x
(a) fix) = )3 cos fix
f"{x) = -0'smpx
f"'(x) = -p^cosfix
/(^' = /S" sin lix
(b) fix) + p^fix) = -/32 sin fix + fi^ism fix) = 0
(c) f ^^\x) = {- ly fi^'' sin fix
/(^--DW = (-!)*+ 1/32*-' cos /Sjc
101. (a) r'ix) = f'{g{x))g'ix)
r'(l)=/'(g(l))g'(l)
Note that ^( 1 ) = 4 and /'(4)
6-2
Also, g'{l) = 0. Thus, r'(l) = 0
(b) s'ix) = g'{fix))f'{x)
s'{4) = g'(f(4))/'(4)
Notethat/(4) = |g'(|
6-4 1 ^
T = -and
6-2 2
/'(4)
Thus, 5'(4)= 1(1
103. g = Vx(x + n)
= ~Jx- + nx
^ = i(;c2 + nx}-'/2{2x + n)
ax 2
2x + n i
Ijx^ + >u
(2;c + «)/2
VaCx + w)
[x + (x + «)]/2
VjcU + «)
£
105. g(x) = |2jc - 3|
2a:- 3
g'{x)
2JC-3
X9t
107. /j(x) = |x|cosx
h '(x) = — \x\ sinx + -r~\ cos x, x ^ 0
Section 2.5 Implicit Differentiation 79
109. (a) f(x) = tan -
/(I) = 1
/W = -sec2 —
.,„ s TT -, TTX TTX TT
/ W=-sec2— -tan— (-
/'(I) = J(2) = f
/"(l)=|(2j(l) = J
/',W=/'(l)(;c-l)+/(l) = |(;c-l)+l.
P.U) = |(f)(x - IF +/'(1)U - 1) +/(!) = |{:c - IP + |(x - 1) + 1
(b)
(c) Pj '5 ^ better approximation than P,
(d) The accuracy worsens as you move away from x = c = 1 .
111. False. Ify = (1 - xY'-, theny' = jd - x)-'/-(-l).
113. True
Section 2.5 Implicit Differentiation
1. x^ + f = 36
2x + 2yy' = 0
^1/2 + y/2 = 9
-1/2
,-1/2
5. .r^ - XV + y' = 4
3x' - jty ' — y + 2yy ' = 0
(2y — x)y ' = y — 3x^
Zy — X
7. x^y - y - X = 0
3.t3y2y'+ 3.A-' - y' - 1 =0
(3;c3>'= - l)v'= 1 - 3xn^
, _ 1 - 3.rK-^
•'■ 3.r3%- - 1
9. x^ - ?>x- + Zxy~ = 12
3x2 _ 3^2y- _ 6^^^, + 4;tyy' + 2r = 0
(4xv - 3x2)y' = 6xy - 3x- - 2y-
j _ 6xy - 3x- — 2y2
^ ~ 4x7 - 3x2
11. sin.v + 2cos 2y = 1
cos X - 4(sin 2y)y ' = 0
4 sin 2v
80 Chapter 2 Differentiation
13. sin.r = ;>:(1 + tany)
cos.r = x{sec'^ y)y' + (1 + tany){\)
, cos X — tan V — 1
y = T"
X sec- y
17. (a) ;c2 + y2 = 16
/ = 16 - a:^
±Vl6 - x^
15. y = sin(xy)
y' = [xy' + >']cos(xy)
y' - X cos{xy)y' = y cos(xy)
y cos(xy)
y =
I — X cos(xy)
(b)
y= Vl6-jc2
y = -V16-x2
/
(c) Explicitly
dy _l
dx
(16 - x?)-'"-(-2x)
(d) Implicitly:
2x + lyy' = 0
Vl6 - ;c2
±716"
19. (a) \(iy-= 144 - 9^;^
^ = 1^044 - 9x2) = ^(,6 _ ^2)
b^yie"
(b)
(c) Explicitly:
J = ±|(16-jc2)-/2(-2x)
3x
-3x -^x
4716 - X- 4(4/3)y 16y
(d) Implicitly:
18x + 32xv' = 0
-9x
16)-
21. xy = 4
xy' + y(l) = 0
xy'= -y
'■ = ?
At (-4,-1): y'=-\
2'- ^"-x= + 4
, , (x2 + 4)(2x) - (x2 -
- 4)(2x)
^•^^ " (x2 + 4)2
^^^
^^^ (x^ + AY
8x
-^ y(x2 + AY
At (2, 0), >>' is undefined.
Section 2.5 Implicit Differentiation 81
25. x2/3 + y2/3 = 5
r-1/3
-1/3
At (8,1): y'= --.
3/^
27. tan(a: + y) = jc
(1 +y')sec2(j( + >) = 1
1 - secHx + y)
sec^U + y)
— tan^U + y)
tan^U + y) + 1
— cin^l
sin^U + y)
x2+ 1
At (0,0): y' = 0.
29.
0
c= + 4)y =
= 8
(^^'
+ 4)y'
+ y(2x) =
y' =
= 0
-2xy
x~ + 4
=
-Zx[S/(x^ +
x^ + 4
-16;c
4)]
(x- + 4)-
. ,^ ,, , -32 1
At(2.1):y' = — =--
Or, you could just solve for y: y = — -
31. (x^ + y2)2 = 4x^
2(x^ + y-){2x + lyy') = 4.r=y' + y(8x)
4j:^ + 4x^ ' + 4xy + 4v^y ' = 4.t^' ' + 8xy
4j:n7 ' + 4y^y ' — 4x^ ' = 8xv — 4.r^ — 4xy^
4y '(j:^ + >'^ ~ X-) = 4(2xv' — .r' — xy~)
At(l, 1): y'= 0.
, _ 2xy — x^ — xy-
x?y + y^ — X-
33.
tany =
y'sec'v =
X
1
.v' =
1
= cos-
•y, -
n
sec- y
2
sec^v =
1 + tan
2y =
1 +
X-
y' =
1
1 +x-
37.
x^
- y^ =
16
Ix -
2yy ' =
y' =
0
x
y
X
- yy' =
0
1 - yy" -
(y'f =
0
\-yy"-
(;)= =
0
T -
- y'y" =
.X-
v"-
1
r -
1
X' _
77
< V < y
35. x^ + y2 = 36
2x + 2vv' = 0
„ y(-l)-^■n■'
y = ; ■■
y-
-V -l--r -
(-f)
-36
^
-^'^ 39. r = .x'
^ 2yv' ' = 3.r-
>\^i
.V"^
<^'
-<
fi
, _ 3£2 _ 3£2 ^ _ 3y .r^ 3y
2y 2y xy 2v \- 2t
„ 2T(3yO - 3y(2)
y =
, ,c
.^^
•^
^
r<
4x^
lx[i • (3y/lv)] - 6y
16
f y3
•1
K^
^ . M
^ 4.1- 4v
4.r=
1>
..^^
•.^>
V\^ ...c
'>
A
.■^'X
82 Chapter 2 Differentiation
41. v^ + v9 = 4
|;c-'/2 + ir'/^' = o
V^
At(9, l),:y'= -J
Tangent line: y - 1 = ~:;(x — 9)
y=-^x + 4
X + Sy - 12 = 0
43. x^ + f = 25
—X
At (4, 3):
Tangent line: y - 3 = —(x - 4) =^ 4;c + 3^ - 25 = 0
Normal line: .v - 3 = -{x - 4) => 3j: - 4>' = 0.
At (-3, 4):
Tangent line: y — 4 = -(x + 3) => 3j: - 4y + 25 = 0
Normal line: y - 4 = -r-ix + 3) => 4;c + 3y = 0.
45. x^ + y^ = r^
2x + 2yy' = 0
y ' = — = slope of tangent line
- slope of normal line
Let (xq, Jq) be a point on the circle. If Xq = 0, then the tangent line is horizontal, the normal line is vertical and, hence, passes
through the origin. H Xg i^ 0, then the equation of the normal line is
yo
y = —x
which passes through the origin.
Section 2.5 Implicit Differentiation 83
47. 25j;2 + I6f + 200;c - \60y + 400 = 0
50.x + 32yy' + 200 - \60y' = 0
, ^ 200 + 50.t
^ ~ 160 - 32.V
Horizontal tangents occur when x = -4:
25(16) + 163'2 + 200(-4) - 160y + 400 = 0
>'(>'- 10) = 0=*>' = 0, 10
Horizontal tangents: (-4, 0), (-4, 10).
Vertical tangents occur when >> = 5:
25jc2 + 400 + 200.t - 800 + 400 = 0
25;c(.x: + 8) = 0 => .v = 0, - 8
Vertical tangents: (0, 5), (- 8. 5).
(-8.5)
49. Find the points of intersection by letting y^ = 4x in the equation 2x^ + y^ = 6.
2^2 + 4x = 6 and (x + 3)ix - 1) = 0
The curves intersect at (1, ±2).
Ellipse:
Parabola:
4x + lyy' = 0
2yy ' = 4
Ix
, 2
y=-j
' ^y
At (1,2), the slopes are:
y'=-l
y'= 1-
At(l, -2), the slopes are:
y'= 1
y'=-l.
Tangents are perpendicular.
51. y = —X and x = sin v
Point of intersection: (0, 0)
y = -x: X = sin y:
y' = -I 1 = y ' cos y
y' = sec y
At (0, 0), the slopes are:
y'= -1 y'= 1.
Tangents are perpendicular.
53. xy= C x^ -f = K
xy' + y = Q 2x-2yy'=Q
y , X
y=-- y=-
. b^ = 4x|
f
f
k
^=.
|lr2 + y2 = 6
C= I
\
'"7-
J-
\
At any point of intersection [x. y) the product of the
slopes is (-y/.ir)(.t/y) = - 1. The curves are orthogonal.
84
Chapter 2 Differentiation
55. 2/ - 3jt* = 0
(a) Ayy' - llr^ = 0
Ayy' = \2x^
' = 12^= 3^
57. cos Try - 3 sin ttj: = 1
(a) — IT sin (Try)^' — 3t7 cos ttat = 0
(b) — 77 sin(T7>')-p - Btt cos(irx)— = 0
at at
, - 3 cos TTX
y = — ■
sin try
59. A function is in explicit form if y is written as a function
of x: y = f(x). For example, y = x^. An implicit equation
is not in tlie form y = f{x). For example, x^ + y^ = 5.
■ / \dy . , Jbi
-smlTT-y)— = 3 cos(Tr;c)— ■
61. (a) x^ = 4(4;c2 _ ^2)
4/ = 16;c2 - y^
f = 4x^ - -x*
4
4x^ - -y^
(b)
v = 3=*9
4;c2 - 7A^
4
36 = 16x' - x^
y^ - ]6x^ + 36 = 0
^,^16± 7256^144^^^^
Note that jc^ = 8 ± 728 = 8 + 2^7 = (l ± Vv)".
Hence, there are four values of x:
-1-V^, 1-V7, -1 + V7, 1 + 7?
x(8 - x^)
To find the slope, 2yy ' = 8x — x^
■y =
2(3)
For a: = - 1 - Jl,y' = jIV? + 7), and the line is
>-, = \[Jl + 7)(x + I + 77) + 3 = ^[(77 + i)x + 877 + 23)].
For X = \ — 77, y' = jl 77 - 7), and the line is
yi = Kv^ - 7)(x - 1 + 77) + 3 = ^[(77 - l)x + 23 - 877].
For j: = - 1 + Jl,y' = -^iVl - l), and the line is
Vi = -1(77 - 7)(;c + 1 - 77) + 3 = -5[(77 - l)x - (23 - 87?)].
Forx = 1 + 77,^' = -5(7^ + 7), and the line is
y, = -}(77 + 7)(;c - 1 - 77) + 3 = -^[{^ + l)x - (877 + 23)].
—CONTINUED—
V
X
-ff
^
Section 2.6 Related Rates 85
61. —CONTINUED—
(c) Equating y^ and y^,
-\{ji - i){x + 1 - v?) + 3 = -|(y7 + i){x - 1 - y?) + 3
(V7 - i){x + 1 - V?) = (77 + 7)(jc - 1 - Vv)
V7;c +V7-7-7JC-7 + 7^7 = Jlx -Jl-l + lx-1- 1 Jl
16V7 = \Ax
877
nx = _ , then v = 5 and the hnes intersect at | „ . 5
63. Let /(a:) = .t^ = ;t^''*, where p and q are nonzero integers and g > 0. First consider the case where p = 1. The derivative of
/(jc) = x'/* is given by
:f [.'/.] = lim f^^^^)-f^-^ = nn, /W^M
dJC :^r^0 ^x t-^x t — X
where t = x + Ajc. Observe that
fit) - f(x) _ f'/" - x^^i _ t'^l" - x^li
t - X
t - X {t"")" - {x^li)"
(fl/9 _ y\lq^(f\-(\lq) + jl-(2/,)^l/<, _!_... + jl/«^I-(2/9) + yA-illqf)
_ 1
~ fl-(l/9) + fl- (2/9)^1/? _|. . . . + jl/9^1-(2/?) + ^1 -(!/«)■
As f — >.i:, the denominator approaches g.x'~"^*'. That is.
Now consider/(.t) = ji^/' = (.x^)'/«. From the Chain Rule,
g iit g q q \ q
Section 2.6 Related Rates
1. y= ^x
dy_ ^ I 1 \dx
dt XlJ'x] dt
dt dt
(a) When .v = 4 and (iv/dr = 3,
dt 2v^^^^~4-
(b) When .r = 25 and dyjdt = 2,
^ = 2725(2) = 20.
dt
3.
.rv' =
4
''1t =
0
dy _
dt
vWv
xjdt
dt
(-
x\dy
\ldt
(a) When x =
= 8.
V = 1
y = 1/2, anddxM= 10.
(b) When .r = 1. v = 4. and dy/dt = -6.
dt
-i,-« = i
86 Chapter 2 Differentiation
5. y = x2 + 1
dx
dt
= 2
dt dt
(a) When ;c = - 1
dy
dt
= 2(-l)(2) = -4 cm/sec.
7.
J =
tan j:
ate _
dt
2
dy
dt
sec^j:
^
A
(a) When x= -it/3,
dy
-j- = (2)2(2) = 8 cm/sec.
dt
(b) When x = 0,
dy
, = 2(0)(2) = 0 cm/sec.
dt
(c) When x = \,
^ = 2(1)(2) = 4 cm/sec.
af
dx dy . .
9. (a) — negative => — positive
(b) — positive => — negative
ar dt
(b) When x = - tt/4,
dy
-j-= (72)2(2) = 4 cm/sec.
(c) When x = 0,
^ = (1)2(2) = 2 cm/sec.
dy
11. Yes, >> changes at a constant rate: ~r = a
No, the rate -p is a multiple of —r-
dt *^ dt
dx
dt'
13. D = Jx-+f = 7x2 + (^2 + 1)2 = 7x^ + 3x2+1
— = 7
dt
dD 1, . , , ,x—i /-,/,,. , \dx 2x^ + 3x
dt 2' 'dt Vx^ + 3x2 +
dx
4x3 + 6x
1 dt Jxf + 3x2+1
15.
A =
77r2
dr _
df
3
dt
2Trr —
dt
(a) When r =
6,
dA _
dt
27r(
(b) When r =
24
dA
dt
= 27r(6)(3) = 36-77 cm2/min.
= 2Tr(24)(3) = 1447rcm2/min.
2 s 2
eh, e
cos X = ~=^ h = s cos -
2 s 2
. 1,, \ L . e\ 6
A = -bh = -\2s sin -II 5 cos -
= yl2sm-cos-l=-sine
dA s2
,de
de 1
(b) -— = — cos fl-— where ^- = - rad/min.
d/ 2 df dt 2
77 dA
^'^^"^=6'd7 = y- 2 8
V3\/l\ 735
^^"^=3'dr=2(2A2
52
(c) If dS/df is constant, dA/dt is proportional to cosft
Section 2.6 Related Rates 87
4 dV
19, V = -7rr3, ^ = 800
3 dt
dV _ , dr
-J- = Airr- — -
dt dt
1 /rfV
i"47^r2UJ"47^r^^^°^^
(a) When r = 30,
"7 = :; — r^Tv^? (800) = — - cm/min.
£?r 47r(30r 9tt
(b) When r = 60, ^
1
dt 47r(60)'
(800)
1
18-n-
cm/min.
21. s = 6;c2
= 3
= 12x
dx
dt
(a)
When
x= \,
ds
— = 12(1)(3) = 36cmVsec.
dt
(b) Whenx = 10,
ds
dt
12(10)(3) = 360cm7sec.
23. V = -irr-h =-Tr(-hAh [since 2r = 3h]
4
— = 10
dt
^ = ^^2 ^ ^ _ 4(rfV/^f)
dr ~ 4 dt"^ dt~ 9-771%-
When/! = 15
d/i _ 4(10)
dt 977(15)- 40577
ft/min.
25.
(a) Total volume of pool = -(2)(12)(6) + (1)(6)(12) = 144 m^
Volume of Im. of water = -(1)(6)(6) = 18 m^
(see similar triangle diagram)
% pool filled = Ts(100%) = 12.5%
(b) Since for Q < h < 2, b = 6/i, you have
V = \bh(6) = 3bh = 3(6h)h = Wr
1 1
dV
... dh
1 dh
36/!— =
= -=>--r
dt
dt
4 dt
1
144/1 144(1) 144
m/min.
88 Chapter 2 Differentiation
27. x2 + / = 252
^ dx dy
Ix — + 2y — = 0
dt dt
dy _ j2£
dt y
dx
dt
Ix . dx
— since —- - 2.
y dt
(a) When x = l,y= 7576 = 24, -f
dt
dy -2(7) -7
24
12
ft/sec.
(b) A = -xy
When;f = IS,)' = 7400 = 20.
rfy
-2(15) _ -3
When;c = 24,j = 7,^ =
dt
dt
2(24)
20
-48
ft/ sec.
ft/sec.
dy dx
iVJt^'Jt.
dx
From part (a) we have a: = 7, 3; = 24, — - = 2,
dt
dy 1
(c)
tan 0 = -
y
^ „ d6 1 dx X dy
sec^ e — = ; r--f
dt y dt
dt
de ^ \,\ dx X dy^
— = cos- 6\- ■ ^ • -r
dt V,y dt y~ dt\
Thus,
dA
dt
)_
2
527
24
12
+ 24(2)
21.96 ftVsec.
Usingx= 7,^ = 24,^ = 2,^
^ dt dt
24
de
12
and cos 8 = —, we have -r = \t7
25
dt
24 \2
25
24^ ^ (24)- 1 12
12
rad/sec.
29. When y = 6, ;t = 712^ - 6- = 6 V3, and
s= Vx- + (12 - y)2
= VIO8 -I- 36 = 12.
;c2 + (12 - y)2 = 52
12 -V
^^
^^
■
n...
X
^^te
y
\
^
y<
n
x-r + {y- \2)-f = s —
dt dt dt
Also. ;c2 -K /= 122
^ dx , ^ dy „ dy —x dx
dt dt dt y dt
™, dx , , ,^J—xdx
Thus, X— -I- [y- 12 — —
dt \ y dt
dx
dt
X +
\2x
ds_
dt'
ds
= 'Jt
dx _ sy ds
(12)(6)
dt \2x ' dt (12)(673)
(-0.2)
5V3
15
m/sec (horizontal)
(iy ^ -;cflDc ^ -6v^ (-VI) ^ J_
rff
y rfr
15
m/sec (vertical).
Section 2.6 Related Rates 89
31.
(a)
s^ =
dx
dt
dy _
dt
-450
-600
dt
dt
^ dt
ds
x{dx/dt)
+ y{dyldt)
100 200
Distance (in miles)
When x= 150 and >> = 200, s = 250 and
ds _ 150(-450) + 200(-600)
dt
250
= -750mph.
250 1
(b) r = — = J hr = 20 min
33.
1
s~ =
X —
dx
dt
90- + x-
30
-28
. ds
2s— =
dt
dx ds X
dt dt s
dx
' dt
Whenj:
s =
= 30,
5oyio
= V90- + 30= =
ds
dt "
^13^'-'^
-28
yio
5 —
8.85 ft/sec
35.
«f
-'' • llv
- 15.x =
6y
y - X
dt
dy 5 dx 5 , _, 25 . ,
^., d{y-x) dv dx 25 ^ 10,.,
90 Chapter 2 Differentiation
37. x(t) = ^sm^,x^- + y^= 1
2 6
277
(a) Period: — 77 = 12 seconds
77/6
(b)When. = i.^7l-(l)^^f,
73'
Lowest point: I 0,
(c) When x = -,y
VMI
and r = 1
39. Since the evaporation rate is proportional to the surface
area, dV/dt = kiAirr-^). However, since V — (4/3)Trr',
we have
dV . ^dr
dt dt
Therefore,
dt dt
dx l/7r\ TTt TT TTt
~r = T 7" COS -T = TX COS —r
dt 2\6/ 6 12 6
X2 + / = 1
^ dx „ dy „ dy -x dx
2x— + 2vT^ = 0=>-f = — .
dt ■ dt dt y dt
Thus,
di
dt
Speed
1/4 TT / 77
-77 / 1 \^ ^ -77 1
■VSt
yi5V12/ 2 24 75 120 •
-7577
120
7577
120
m/sec
41.
py\.^
1.3^V0.3^+V'3^ = 0
dt dt
^ dt dt
..3pf=-v4
dt
dt
43.
tan e =
30
3 m/sec.
sec- d
dd_ ]_di
dt ~ 30 dt
de 1 ^dy
— - = TT COS' e • — -
rfr 30 rff
When >- = 30, e = 77/4 and cos 6 = v z/2. Thus,
rfe 1 /l\,,v 1 .,
^=30l2r^ = M'^/^^'=-
Section 2.6 Related Rates 91
45.
tan e = -, y = 5
X
^= -600mi/hr
dt
.^li
,-T
,=5
, ^^.dO 5 dx
(sec^e)— = — r • —
do
dt ''''\-7^)-d-t = U[-7^)jt
= (-§ )(|)f = (-sin2 0)(|)(-6OO) = 120sin'9
(a) When 6 = 30°, ^ = ^ = 30 rad/hr = ^ rad/min.
at 4 2
(b) When e = 60°, y = 12o(|) = 90 rad/hr = | rad/min.
(c) When 6 = 75°, ^ = 120 sin^ 75° « 1 1 1.96 rad/hr = 1.87 rad/min.
dt
47, — = (10rev/sec)(2irrad/rev) = 20'n-rad/sec
(a) cos 9 = —
. „de I dx
-'"^^Yt^ToTt
^=-30sin.^ =
dt dt
(b) 2
wo
A A
0
vAT
-30 sin e(207r) = -60077 sin e
(c) \dx/dt\ = I -60077 sin 0| is greatest when sin e = 1 => 6= (tt/I) + mr(oT90° + n ■ 180°)
\dx/dt\ is least when 6 = mriorn • 180°).
(d) For e = 30°,
dx
dt
-600T7sin(30°) = -6OO77- = - 3OO77 cm/sec.
For e = 60°, ^ = -60077sin(60°) = -60077^ = -300V^ 77 cm/ sec
dt 2
49. tan e = — => j: = 50 tan e
— - = 50 sec- 6 —-
dt dt
2 = 50 sec-
1^
— - = rrcos^ e, -— < e < —
dt 25 4 4
92 Chapter 2 Differentiation
51. x~ + )r = 25; acceleration of the top of the ladder = -jj
dx dy
First derivative: lx—- + 2y— = Q
dt ■ dt
dx , dy
X— + y— = 0
dt dt
.... d-x , dx dx , dhi dy dy .
Second denvative: x — -r + —r ' —r ~^ TT^ + ~; T = 0
dt- dt dt dt^ dt dt
d^ (I
dt^ '
dy^ _1_
dt 12'
dx
dt
dh _ fdx\2 _ fdyy
'^ \dt) [dtj .
dx . . d^x
When X = 7, y = 24, — - = — — -, and — - = 2 (see Exercise 27). Since -— is constant, -rr = 0.
■^ -'* '^ -'• dt dt^
d^ ^ 1
dt'^ " 24
-7(0) - {ly
24
:] =
1441 24
625
144
= -0.1808 ft/sec^
53. (a) Using a graphing utility, you obtain mis) = -0.8815^ + 29.105 - 206.2
(b)^ = ^j;=(- 1.762. + 29.10)f
dt ds dt dt
ds
(c) Ifr = s(1995J, then. = 15.5 and— = 1.2.
Thus. ^ = (- 1.762(15.5) + 29.10)(1.2) = 2.15 million.
dt
Review Exercises for Chapter 2
1. fix)
= X- -Ix + Z
f'(x)
f(x + Ax)-f{x)
= lim
Aj->o Ax
,. [{x + Axf - 2{x + Ax) + 3]- [x^ - 2x + 3]
Ax-»0 Ax
•
,. ix" + 2x(Ax) + {Ax)- -2x- 2(Ax) + 3) - {x^ -
- 2x + 3)
iit->o Ax
,. 2x(Ax) + (Ax)- - 2(Ax) ,. ,^
= hm — ^^ — ^--~ 5^ — - = lim (2a: + A;c - 2) =
AJ-.0 Ax _ Ax-^O
= 2x - 2
3. f(x)
= Vx + I
5. /isd
fix)
fix + Ax) -fix)
— lim
^x-M Ax
,. (Vx + Ax + 1) - (Vx + 1)
5. / is differentiable for all .t =?^ — 1 .
= lim
Ax->0
lim
Ax
^x + Ax - yx Vx + Ax + Vx
Ax Vx + Ax+ 7x
(x + Ax) — X
Ax-^o Ax(Vx + Ax + Vx)
r 1 1
^-0 Vx + Ax + Vx 2Vx
Review Exercises for Chapter 2 93
7. fix) = 4 - |;c - 2|
(a) Continuous at x = 2.
(b) Not differentiable at jr = 2 because of the sharp turn
in the graph.
4 1
9. Using the limit definition, you obtain g '{x) = -x - -
4 1 -3
Atx=-l,g'(-l)=---- = —
11. (a) Using the limit defintion, f\x) = "ix^.
At X = - !,/'(- 1) = 3. The tangent line is
>-- (-2) = 3(;c -(-!))
.V = 3x + 1
(b) -4f
13. g'(2) = lim.
gW - g(2)
2 X-1
= lim-^(--V'
jr->2 JT — 2
,. A^ - ;c2 - 4
= lim
x-*^ X — 2
,. (x - 2)U2 + x + 2)
= lim r
lim U^ + ;c + 2) = 8
a:-»2
IS.
17. >- = 25
y' = 0
19. /W = x»
f'(x) = ar'
21. /i(r) = 3r*
A '(f) = 12f'
23. /U) = .r3 - 3x'
f'(x) = 3x- - 6x = 3.x(.r - 2)
25. A(x) = 6V^ + 3^ = dr'/2 + Sx'/'
/I'W = 3;t-'/2 + X-V3 = 3 _^ 1
29. fid) = 20 - 3 sin 9
/'(0) = 2 - 3 cos e
27. g(r) = |r-
-4 -4
31. fie) = 3 cos e
sin e
fie) = -3 sine -
cos e
94 Chapter 2 Dijferentiation
33. F = 200yr
100
F'{t) =
Vr
(a) When T = 4, F'(4) = 50 vibrations/sec/lb.
(b) When 7=9, F'(9) = 335 vibrations/sec/lb.
35. s(t) = - 16f2 + 5o
^(9.2) = - 16(9.2)2 + So = 0
5o = 1354.24
The building is approximately 1354 feet high (or 415 m).
37. (a)
20 40 60
Total horizontal distance: 50
(b) 0 = X - 0.02x2
0=x 1
50
implies x = 50.
39. x{t) = f - 2t + 2 = it- 2){t - 1)
(a) vit) = x'it) = 2t - 3
ait) = v'(r) = 2
(c) v(f) = Oforr = i
^ = (l-2)(|-i) = (4)(l) = -i
(c) Ball reaches maximum height when x = 25.
(d) y = x- 0.02x2
y'= 1 - 0.04x
y'(0) = 1
y'ilO) = 0.6
^'(25) = 0
y'i30) = -0.2
y'(50)=-l
(e) y'(25) = 0
(b) v(f) < 0 for f < 5.
(d) x(f) = Oforf= 1,2.
lv(l)| = |2(l)-3| = l
|v(2)| = |2(2) - 3| = 1
The speed is 1 when the position is 0.
41. fix) = (3.x2 + 7)(x2 - 2.x + 3)
fix) = (3x2 + 7)(2;^ - 2) + (x2 - 2x + 3)(6x:)
= 2(6^3 - 9x2 + 16^ _ 7)
45. fix) = 2x - x-2
fix) = 2 + 2x-^ = 2( 1 + -J
. _ 2(x3 + 1)
X3
43. hix) = Jlc sin ;t = x'/2 sin x
1
h'ix) =
47. /(:c) =
fix) =
2V5"
X2 +
+ J~XI.
1
Jt2- 1
(x2 - l)(2x + 1) - (^2 + X - l)(2x:)
ix" - 1)2
-(;c2+l)
(x2 - 1)2
49. fix) = (4 - 3x2)-'
6x
/'(x) = -(4 - 3x2)-2(-6x) ^4 _ 3^,^,
53. >> = 3^2 sec X
y'
ji. y —
cosx
cos X ilx) - x\- sin x)
cos2 X
2x cos X + x2 sin x
cos2x
55. y = -jctanjc
3; ' = — x: sec2 X — tan X
Review Exercises for Chapter 2 95
57. y = X cos j: — sin j:
y' = — j: sin ;c + cos X - cos j: = -xsin;c
59. gU) = r^ - 3r + 2
g'it) = 3^2-3
g"{t) = 6r
61. f(0) = 3 tan 9
f'(d) = 3 sec2 6
f'(e) = 6 sec e (sec 6 tan 0) = 6 sec^ 0 tan
63. >> = 2 stn j: + 3 cos x
y' = 2 cos x — 3 sin x
y" = — 2 sin ;t — 3 cos x
y" + y = — (2 sin ;c + 3 cos x) + (2 sin jc + 3 cos x)
= 0
65. f(x) = (1 - .r3)'/2
1,
/'W=^l-x5)->/2(-3x2)
3;c2
2Vl -:t3
67. /zW = (^ ^^'
h'(x) = 2
X^ + ly
;c-3\/y-+ 1)(1) - (;c - 3)(2x)\
a:= + 1
+ 1)^ ;
2U - 3)(-.t^ + fa + 1)
69. /(i) = {s- - 1)5/2(53 + 5)
f'(s) = (52 - 1)V2(3^2) + (^ + 5)(|)(^2 _ 1)3/2(2^)
= sis' - l)3/2[35(i2 - 1) + Sis' + 5)]
= sis' - 1)3/2(8^3 - 3i + 25)
71. y = 3 cos(3.r + 1)
y' = -9sin(3;c + 1)
73. y = "z CSC 2x
y ' = r{ — CSC 2x cot 2i:)(2)
- CSC 2x cot 2t
.r sin 2x
= r{l — cos 2x) = sin^.r
2 2
77. y = - sin3/2 x - - sin'^/^x
= sinl/2
inV2 ,
sm'''^ JTCOS j: — sm^'-xcos j:
(cos j:)Vsin j:(1 — sin^x)
c3
= ( cos= xl V sin a:
79. V =
.T + 2
, _ ix + 2)Trcos TT.r — sin tt.v
•'' " (-v + 2)2
81. /(/) = fit - ly
fit) = t(t- DVt - 2)
The zeros of/' correspond to the points on the graph off
where the tangent line is horizontal.
83. gix) = 2x(.r + D-'/^
X + 2
g'U) =
ix + 1)3/2
g ' does not equal zero for any value of t in the domain.
The graph of g has no horizontal tangent lines.
•■\
^
" J
96 Chapter 2 Differentiation
85. fit) = {t+ l)i/2(r + l)'/3 = (r + l)V6
5
/'(f)
6(f + !)'/«
/' does not equal zero for any x in the domain. The
graph of/has no horizontal tangent lines.
^
^
/■
87. y = tanVl - x
c=yr
2Vl -.«
y ' does not equal zero for any x in the domain. The graph
has no horizontal tangent lines.
r^
89. y = 2;t2 + sin 2jc
>-' = 4;c + 2cos2i
y" = 4 — 4s'm2x
91. /(a:) = cot X
93. /(r) =
fix) = -csc^a;
/"= -2cscjf(-cscj:
cotjc)
/'(f) =
= 2 csc^ X cot ;c
/"(f) =
t
(1-
ty
f +
1
(1-
f)'
2(f4
-2)
(1 - f)^
95. gie) = tan 30 - sin(e - 1)
g'ie) = Ssec^se- cos(e- i)
g"(e) = 18 sec2 3etan 361 + sin(e - 1)
97. r= 700(f2 + 4r+ 10)-'
- 1400(f + 2)
" (f2 + 4r + 10)2
(a) When f = 1,
^, -1400(1 + 2)
(1 + 4 + 10)2
(c) When ; = 5,
- 1400(5 + 2)
== - 18.667 deg/hr.
r
(25 + 30 + 10)
5= -3.240 deg/hr.
(b) When r = 3,
^, _ - 1400(3 + 2)
(9 + 12 + 10)2
(d) When t = W,
-1400(10 + 2)
7"
(100 + 40 + 10)2
-7.284 deg/hr.
= -0.747 deg/hr.
99. x2 + 3xy -f- y = 10
2x -I- 3xy' -I- 3y -I- 3>'2y'= 0
3(a: + /)>>'= -(Ix -(- 3>')
-(2A: + 3y)
^ Hx + f)
101.
yv^t ~ x^ = 16
2V^ — X , __ 2vxy — y
iVy ^ " 2Jx
, ^ ijxy - y _ ijy ^ lyjx - yVy
2jx iVxy — X Ixjy — x^x
Review Exercises for Chapter 2 97
103. xsxny = y cos x
{xcosy)y' + smy = ->> sinj: + >''cos;c
>''(^cos>' - cos x) = — ysinjt — siny
, _ y sin a: + sin y
cos X — X cos y
105. Jt^ + y2 = 20
2x + lyy' =Q
At (2, 4): y'= --
i4
\,
Tangent line: y — 4 = — xU - 2)
;c + 2y - 10 = 0
Normal line; y - 4 = l{x - 2)
2x-y = Q
107. y= ^x
dy
dt
= 2 units/sec
dy 1 ^v ^^2^^ = 4^
dt 2jx dt
dt
dt
1 dx
(a) When x = — , — = 2 V2 units/sec.
dx
(b) When j: = 1, — = 4 units/sec.
dt
dx
(c) When j: = 4, — = 8 units/sec.
dt
109.
5
h ~
1/2
2
S =
^'
dt
1
Width of water at depth h:
w = 2 + 2s = 2 + 2\hx
A + h
''-f(^-^> = j«*'*
dV 5,, ,.dh
dt
dh ^ 2{dV/dt)
dt ~ 5(4 + h)
dh
When h = 1. ^ = rr m/min.
dt 25
111. s{t) = 60 - 4.9r
s\t) = -9.8r
5 = 35 = 60 - 4.9f-
4.9r= = 25
5
tan 30°
74.9
.r(f) = Jls(t)
^=73^=V3(-9.8h^
= - 38.34 m/sec
98 Chapter 2 Differentiation
Problem Solving for Chapter 2
1. (a) x~ + (y - r)^ = r^ Circle
X- = y Parabola
Substituting,
(■y _ ;.)2 = ^ _ y
y'^ — 2ry + r' = r^ — >>
y^ — 2ry + >> = 0
y{y - 2r + 1) = 0
Since you want only one solution, let 1 — 2r = 0 => r
Graph y = x'^ and x'^ + [y - 5)' = 4
(b) Let (x, y) be a point of tangency: x^ + (y - b)^ = 1 => 2j: + llj" - b)y' = 0 => y '
y = ;c'^ => >; ' = 2x (parabola). Equating,
r
2jc^
Z.-y
(circle).
b-y
2{b - y) = 1
Also, a;^ + (y - Z))^ = 1 and y = x^ imply
y + (y - fc)2 = 1 =* y + l^y - (y + ^j j = 1
Center: 0
Graph y = x- and x-
■ y--= l=>y = - and/7 = -.
3. (a) f(x) = cos;c
/(O) = 1
/'(0) = 0
P,{x) = 1
(c)
P\ix) = % + OiX
P,(0) = ao=»ao= 1
P'i(O) = a, =* a, = 0
(b) f(x) = cosx
/(O) = 1
/'(O) = 0
/"(0)=-l
P^ix) = 1 - I;c2
j:
-1.0
-0.1
-0.001
0
0.001
0.1
1.0
cosx
0.5403
0.9950
« 1
1
« 1
0.9950
0.5403
i'aW
0.5
0.9950
= 1
1
" 1
0.9950
0.5
PjW is a good approximation of f{x) = cos x when x is near 0.
(d) fix) = sin;c
/(O) = 0
/'(O) = 1
/"(O) = 0
/"'(0)= -1
P^ix) =x-\x^
Oq + a,j; + flj^^ + ^s-"^
P,m= ao^ao = 0
P\{0) = a, =^ a, = 1
^"3(0) = 2^2 => 02 = 0
^"'3(0) = 603 => 03 = -
PjW = Oq + OiX + a^K-
P^iO) = flo =5> ao = 1
/"2(0) = ai=»a, = 0
P"2(0) = 2a2^a2= -i
Problem Solving for Chapter 2
99
5. Letp(jc) = Aj^ + Bx^ + Cx + D
p'{x) = 3Ax- + 2Bx + C
At (1, 1); A + B + C + D = 1 Equation 1
3A + 2B + C =14 Equation 2
At (-1, -3): -A + B-C + D=-3 Equations
3A - 2B + C = -2 Equation 4
Adding Equations 1 and 3: 25 + 2D = -2
Subtracting Equations 1 and 3: 2A + 2C = 4
Adding Equations 2 and 4: 6A + 2C = 12
Subtracting Equations 2 and 4: 45 = 16
Hence, S = 4 and £> = k(-2 - 2B) = -5
Subtracting 2A + 2C = 4 and 6A + 2C = 12, you obtain 4A
Thus, pW = 2^3 + 4x- - 5.
■ A = 2. Finally, C = 5(4 - 2A) = 0
7. (a) .x^ = a-x^ - a^y^
a^y- = a-x — x^
±Va-.v--.V>
„ . Va-.r- - -r^
Graph: >>, = and Vj =
^cP-x- — .r*
(b)
frb
^
a= 1
vj.,
(±a, 0) are the Ar-intercepts, along with (0, 0).
(c) Differentiating implicitly,
Ax^ = 2a- .1: — 2a^yy'
y = — r-^; = r = Q^2x- = a-=^x = —7=.
2ay a^ V2
4
->
a-
V = ±7
^ . I a a\ I a a\ ( a a\ l-a a\
Four poutts: ^-^, -j. ^^, --j. ^-^. ^ j. (yj' "^J
100 Chapter 2 Differentiation
9. (a)
(b)
90 100
jVo/ drawn to scale
(0, 30)
1.6)
no. 3)
60 70
!^ot drawn to scale
Line determined by (0, 30) and (90, 6):
y-30
30-6
0-90
U-0)
24
'90-" 15^^
?
+ 30
-4 10
When X = 100, y = -j-r(100) + 30 = — > 3 => Shadow determined by man.
Line determined by (0, 30) and (60, 6):
-2
When x = 70,y = -z-(70) + 30 = 2 < 3 => Shadow determined by child.
(c) Need (0, 30), {d, 6), {d + 10, 3) collinear.
30-6
6-3 24 3
0' d
d- id+ 10)^ d 10
d = 80 feet
dx
(d) Let V be the length of the street light to the tip of the shadow. We know that — =
dt
For X > 80, the shadow is determined by the man.
y y ~ X 5 , dy 5 dx —25
5^ = -^=^^ = ?^'^^ = 4^ = ~-
For X < 80, the shadow is determined by the child.
y_ ^ y - X
30 3
Therefore,
10
10 100 _,rfy 10^ -50
dt
-25
4
-50
9
;c > 80
0 < .x < 80
dy
—r is not continuous at j: = 80.
dt
11. L'(x) = lim
Ajt->0
= lim
Aj:->0
L{x + ^x) - L{x)
Ax
Ljx) + L(Ax) - L(x)
Ax
lim
L{Ax)
<ii^o Ajc
Also, Z.'(0) = lim
UAx) - L(0)
Ax->0 Ax
But, L(0) = 0 because Z,(0) = L(0 -I- 0) = L(0) -f- L(0) => L(0) = 0.
Thus, I, W = Z.'(0), forall;c.
The graph of L is a line through the origin of slope L'{0).
Problem Solving for Chapter 2 101
13. (a)
z (degrees)
0.1
0.01
0.0001
sinz
0.0174524
0.0174533
0.0174533
(b) lim^-^- 0.0174533
z->0 z
, . ^ ,. smz 77
In fact, lim —
J->0 z
(c) —(sin z) = lim
dz A;-»0
= lim
180
sin (z + Az) - sin z
Az
sin z • cos Az + sin Az • cos z - sin z
Az
I sin z r + lim
oL V Az /J Aj:-»oL
= lim I sin z
= sinz(0) + cosz(j^ =-f^cosz
cos z
sin Az
Az
(d) 5(90) = sin(^ 9o) = sin ^ = 1; C(180) = cos(^ 180 ) = - 1
—S{z) = — sin(cz) = c • cos(cz) = 757:C(z)
az az ISO
(e) The formulas for the derivatives are more complicated in degrees.
15. jit) = a 'it)
(a) j(t) is the rate of change of the acceleration.
(b) From Exercise 102 in Section 2.3,
s{t) = -8.25f2 + 66;
v(r) = - 16.5f + 66
a(t) = - 16.5
a'(t)=j(t) = 0
CHAPTER 3
Applications of Differentiation
Section 3.1 Extrema on an Interval 103
Section 3.2 Rolle's Theorem and the Mean Value Theorem . 107
Section 3.3 Increasing and Decreasing Functions and
the First Derivative Test 113
Section 3.4 Concavity and the Second Derivative Test .... 121
Section 3.5 Limits at Infinity 129
Section 3.6 A Summary of Curve Sketching 136
Section 3.7 Optimization Problems 145
Section 3.8 Newton's Method 155
Section 3.9 Differentials 160
Review Exercises 163
Problem Solving 172
CHAPTER 3
Applications of Differentiation
Section 3.1 Extrema on an Interval
Solutions to Odd-Numbered Exercises
1. fix)
fix)
x^ + 4
(^ + 4)(2;c) - U=)(2t)
&x
U2 + 4)-
tr- + 4y
/'(O) = 0
3. fix) =x +
27
2x2
27
X + —x
fix) = 1 - 27x-3 = 1
27
/'(3) = 1 - |J = 1 - 1 = 0
5. /(x) = U + 2)2/3
f'i—2) is undefined.
7. Critical numbers: x = 2
X = 2: absolute maximum
9. Critical numbers: x = 1, 2, 3
X = 1,3: absolute maximum
X = 2: absolute minimum
11. f(x) = xKx -3) = x^ -3x-
f'(x) = 3;c2 - 6x = 3x(.t - 2)
Critical numbers: x = 0, x = 2
13. ^(f) = fv/4 - t, t < 3
gV) = r
i(4_ f)-l/2(_l)| + (4-,)l/2
= ^4-r)-'/2[-f + 2(4-r)]
8 - 3r
2V4 - /
Critical number is t
15. hix) = sin^x + cosx, 0 < .t < 27r
h'(x) = 2 sin .t cos AT — sinx = sinx(2 cos.r — 1)
TT 5 TT
On (0, 27r), critical numbers: x = — , x = tt, x = —
17. fix) = 2(3 - x), [- 1. 2]
fix) = - 2 => No critical numbers
Left endpwint: (- 1, 8) Maximum
Right endpoint: (2, 2) Minimum
19. fix) = -X- + 3x. [0. 3]
/'(x)= -2r + 3
Left endpoint: (0. 0) Minimum
Critical number: (2.4) Maximum
Right endpoint: (3. 0) Minimum
103
104 Chapter 3 Applications of Differentiation
21. f{x)=x^-jx', [-1,2]
f'(x) = 3x2 - 3x = 2x(x - 1)
Left endpoint: \~'^'~2J '^™'""'"
Right endpoint: (2, 2) Maximum
Critical number: (0, 0)
V
Critical number: 1,
23. fix) = 3a;2/3 - 2x, [- 1, 1]
/'W = 2;c-'/3 - 2 = 2(LzJ^
Left endpoint: (- 1, 5) Maximum
Critical number; (0, 0) Minimum
Right endpoint: (1, 1)
25. ,W = -^,[-l,l]
1
6t
(t^ + 3)2
g'it)
Left endpoint: I ~ 1. T 1 Maximum
Critical number: (0, 0) Minimum
Right endpoint: ( 1. T ) Maximum
27. his) = y-h;, [0, 1]
h'is) =
s-2'
-1
(^ - 2)2
Left endpoint: I 0, - — 1 Maximum
Right endpoint: (1, - 1) Minimum
29. fix) = cos TTX, 0, -
fix) = - TT sin TTX
Left endpoint: (0,
1) Maximum
/l V3^
Right endpoint: I -, — r-
Minimum
4 7131: r, «n
31. y = - + tan—, [1, 2]
X O
, -4 77 ^ TTX _
TT , ira 4
On the interval [1,2], this equation has no solutions.
Thus, there are no critical numbers.
Left endpoint: (l, v^ + 3) = (1, 4.4142) Maximum
Right endpoint: (2, 3) Minimum
33. (a) Minimum: (0, —3)
Maximum: (2, 1)
(b) Minimum: (0, -3)
(c) Maximum: (2, 1)
(d) No extrema
35. fix) = x^-2x
(a) Minimum: (1, — 1)
Maximum: (—1,3)
(b) Maximum: (3, 3)
(c) Minimum: (1, —1)
(d) Minimum: (1, — 1)
Section 3.1 Extrema on an Interval 105
37. f(x)
2x + 2, 0 < .r < 1
4x-, 1 < X < 3
Left endpoint: (0, 2) Minimum
Right endpoint: (3, 36) Maximum
39. fix) = J— ^A\ A]
Right endpoint: (4, 1) Minimum
41. (a) 5
(1.4.7)1
(0.4398.-1.0613)
Maximum: (1, 4.7) (endpoint)
Minimum: (0.4398, - 1.0613)
(b)
fix) = 3.2^5 + 5jc3 - 3.5x, [0, I]
fix) = 16.r^ + 15^2 - 3.5
16x^ + \5x- - 3.5 = 0
, ^ -15 ± Jil5)- - 4i\6)i-3S)
2(16)
-15 ± 7449
32
/- 15 + .
V 32
'449
- 0.4398
/(O) = 0
/( 1 ) = 4.7 Maximum (endpoint)
<V^
/449
= -1.0613
Minimum: (0.4398, - 1.0613)
43. fix) = (1 + x3)'/2, [0, 2]
fix) = ^Ki + x^)-''-
f%x) = J(X* + 4x)(l + .t5)-3/2
f"\x) = -^(.t« + 20x^ - 8)(1 + x3)-V2
O
Setting/'" = 0, we have .r« + 20.r3 -8 = 0.
jc3 =
-20 ± 7400 - 4(1)F8)
x= i/- 10 ± yiM = 73 - 1
In the interval [0, 2], choose
= V- 10 ± v-^ = 73 - 1 = 0.732.
/"( {/-\0 + 7i08) = 1.47 is the maximum value.
45. fix) = ix+ l)-^\ [0, 2]
/'(x)=|(x+l)-'/3
fix) = -^(.T + i)-*/3
f"{x) = ^(.v + 1)-'/^
/(4)(.^) = _|^(^ + 1
-10/3
/^%) = f§u + l)-"/^
|/*^*(0)| = — is the maximum value.
81
106 Chapters Applications of Differentiation
47. f(x) = tan X
/is continuous on [0, 7r/4] but not on [0, tt]. iim tan j: = oo.
X — »7r/2
49.
51. (a) Yes
(b) No
53. (a) No
(b) Yes
55. P = VI - RI^ = 12/ - 0.5/2, 0 < / < 15
P = 0 when / = 0.
P= 67.5 when /= 15.
P'= 12 - /= 0
Critical number: / = 12 amps
When / = 12 amps, P = 72, the maximum output.
No, a 20-amp fuse would not increase the power output.
P is decreasing for / > 12.
57.
3s^l^-C0Sd\ TT „ IT
s = 6hs + — ■ - — I- < e<-
2 \ smO / 6 2
dS^3fi
dd 2^
• Vscsc e cot 0 + csc^ e)
= -^csc g{- Vscot e + esc e) = 0
CSC 6 = V^cot 0
sec S = Vs
e = arcsecVs = 0.9553 radians
5|'f) = 6fo + ^V3)
5(arcseCv/3) = 6fc + -~-{j2)
S is minimum when d = arcsecVS = 0.9553 radians.
59. (a) y = ax^ + bx + c
y' = 2ax + b •• ^
The coordinates of S are (500, 30), and those of A are (- 500, 45).
From the slopes at A and B,
-1000a + b= -0.09
1000a + b = 0.06.
Solving these two equations, you obtain a = 3/40000 and fc = -3/200. From the points (500, 30) and (-500, 45),
you obtain
30 =
45 =
40000
3
50°' + '"^[wo^ ^ '^
40000 50°' -^ii)^^-
75
In both cases, c = 18.75 = — -. Thus,
4
3 2__3_ 75
^ 40000"^ 200^ 4'
—CONTINUED—
Section 3.2 Rolle's Theorem and the Mean Value Theorem 107
59. —CONTINUED—
(b)
X
-500
-400
-300
-200
-100
0
100
200
300
400
500
d
0
.75
3
6.75
12
18.75
12
6.75
3
.75
0
For -500 < X <Q,d={a:^ + bx + c)- (-0.09^).
For 0 < ;c < 500, rf = (or^ + to + c) - (0.06.r).
(c) The lowest point on the highway is (100, 18), which is not directly over the point where the two hillsides come together.
61. True. See Exercise 25.
63. True.
Section 3.2 Rolle's Theorem and the Mean Value Theorem
1. Rolle's Theorem does not apply to /(j:) = 1 — |x - 1|
over [0, 2] since/ is not differentiable at .r = 1.
5. fix) = xJITa
j:-intercepts: (-4. 0), (0, 0)
f'{x) =x^x + 4)-i/2 + {x + 4)1/2
)-'/^(f
= {x + A)-"A^+ (x + A)
fix) = I 2^ + 4 ]{x + 4)-'/2 = 0 at;c =
3. fix) =x^ - x-2 = ix-2)ix+ I)
jc-intercepts: (-1,0), (2,0)
fix) = 2x-l = 0atx = -.
7. fix) = x--2x, [0, 2]
/(0)=/(2) = 0
/is continuous on [0, 2]. /is differentiable on (0, 2).
Rolle's Theorem applies.
fix) = 2x - 2
2x-2 = 0^.t=l
c value: 1
9./(;c) = (:c- l);.r-2)(x-3),[l,3]
/(l)=/(3) = 0
/is continuous on [1, 3]. /is differentiable on (1, 3).
Rolle's Theorem applies.
fix) =x^ - 6x- + ll.r - 6
fix) = 3x- - \2x+ 11
3a- - 12x + 11 = 0=>x= ~
11. /(.r)=x2/3- l,[-8,8]
/(-8)=/(8) = 3
/is continuous on [—8. 8]. /is not differentiable on
(—8, 8) since /'(O) does not exist. Rolle's Theorem does
not apply.
6-73 6 + v^
108 Chapters Applications of Differentiation
/(-l)=/(3) = 0
/is continuous on [- 1, 3]. (Note: The discontinuity, x = —2, is not in the interval.)/is differentiable on (— 1, 3). Rolle's
Theorem applies.
f,, X (x + 2)(2;c - 2) - U^ - 2x - 3)(1) .
/ U) = 7 — -^:^ = 0
ix + 2)2
c value: — 2+V5
x^ + 4;c - 1
ix + ly
= zi4M = _2±V5
15. /W = sinx, [0,2it]
/(0)=/(27r) = 0
/is continuous on [0, 2Tr]./is differentiable on (0, lir).
Rolle's Theorem applies.
f'{x) = cos X
77 377
c values: — , — —
2 2 V
17. fix) = — - 4 sin^jc,
■ [»f]
/(o)-/m.o
/is continuous on [0, 77/6]. /is differentiable on (0, 77/6).
Rolle's Theorem applies.
fix) = 8 sin a: cos x = 0
77
19. fix) = tan X, [0, 77] ,
/(0)=/(77) = 0
/is not continuous on [0, 77] since/(T7/2) does not exist.
Rolle's Theorem does not apply.
77
8 sm X cos X
477
1 • 0
-smlx
3
277
sin2x
1
— arcsm
.217/
x
X =»
0.2489
c value:
0.2489
.1. fix) =
= w-
1, [-1,1]
/{-1) =
= /(!) =
0
/is continuous on [- 1, l]./is not differentiable on
(— 1, 1) since/'(0) does not exist. Rolle's Theorem does
not apply.
Section 3.2 Rolle's Theorem and the Mean Value Theorem 109
23. fix) = 4a: - tan ttx.
'4' 4
A-4J^A4J-o
/is continuous on [-1/4, 1/4]. /is differentiable on
(- 1/4, 1/4). Rolle's Theorem applies.
f'{x) = 4 - V sec- TTX = 0
25. fit) = -16f2 + 48r + 32
(a) /(I) =/(2) = 64
Q^) V = f'(t) must be 0 at some time in (1 , 2).
/'(f) = -32r + 48 = 0
t = - seconds
sec* TTX = —
77
sec TTX = ±-
'77
^1 1 \ Jtt
X = ±— arcsec — ■;= = +— arccos -^^ —
""■ V77 •"■ 2
« ±0.1533 radian
c values: ±0.1533 radian
27.
tangent line secant line
29. fix) = ^^, [0, 6]
/has a discontinuity at jr = 3.
31. fix) = x~ is continuous on [-2, 1] and differentiable on
(-2,1).
/(l)-/(-2) 1-4
1 - (-2) 3
= -1
fix) = It = - 1 when x = --. Therefore.
'=-2-
33. fix) = .r-'^ is continuous on [0, 1] and differentiable on
(0, 1).
/(l)-/(0)
1 - 0
= 1
fix) = j.v-'/5 = 1
27
110 Chapters Applications of Differentiation
35. fix) = V2 - X is continuous on [— 7, 2] and
differentiable on (—7, 2).
/(2)-/(-7)_0-3 _ _1
2 - (-7)
f'(x) =
9
3
-1
1
272-
X
3
272-
X =
= 3
72-
X -
3
" 2
2 -
X -
_ 9
" 4
X =
1
4
37. /(x) = sin X is continuous on [0, -ir] and differentiable on
(0, tt).
/(7r)-/(0) 0-0
= 0
77-0 V
f'(x) = cos X = 0
77
/w = -
X
yon
1
2'
2
X +
(a)
1
:^
^,,v^^secant
-1
(b) Secant line:
Slope =^(^^^
-/(-1/2)
-(-1/2)
2/3 -(-1)
5/2
2
3
2
= f(;c-2)
3.y - 2 =
= 2x - 4 •
,
Sy-
- Ix
f 2 =
= 0
(c) f'{x) =
(x + 1)^ 3
{x + 1)^ = 1
j: = -1 ± ^
V 2
-..f
In the interval [- 1/2, 2], c = - 1 + (76/2).
-1 + (76/2) _ -2 + 76 -2
f{c)
[-1 +(76/2)] + 1
2
76 76
+ 1
1l /f\
Tangent line; j - 1 + -^ = -Ax + 1
, ^ 76 2 76_^2
y — 1 -I — :::— = -x — - — i- -
^ 3 3 3 3
33- - 2;c - 5 + 276 = 0
Section 3.2 Rolle's Theorem and the Mean Value Theorem
111
41. /W = v^, [1,9]
(1,1), (9. 3)
3-1 I
1
(a)
(b) Secant line: >> — 1 = -(x - 1)
1 ^3
y = ~x + —
4 4
0 = X - 4^ + 3
(c)
fix)
ijk
/(9)-/(l) 1
9-1 4
1 _ 1
2Vc"4
7? = 2
c = 4
{c,/(c)) = (4, 2)
'^ = /'(4) = 4
Tangent line: y - 2 = -(j: - 4)
V = -.X + I
4
0 = .t - 4>' + 4
43. 5(f) = -4.9f2 + 500
5(3) - 5(0) _ 455.9 - 500
(a) K,
3-0
14.7 m/sec
(b) 5(f) is continuous on [0, 3] and differentiable on (0, 3).
Therefore, the Mean Value Theorem applies.
v(f) = s\i) = -9.8r = - 14.7 m/sec
-14.7
-9.8
1 .5 seconds
45. No. Let/(.r) = .r= on [- 1, 2].
/'(.t) = Ix
/'(O) = 0 and zero is in the inter\al (—1,2) but
/(-l)^/(2).
47. Let 5(t) be the position function of the plane. If f = 0 corresponds to 2 p.m., 5(0) = 0, 5(5.5) = 2500 and the Mean Value
Theorem says that there exists a time fg, 0 < fg < 5.5, such that
S%) = v(ro) = ^ffy » 454.54.
Applying the Intermediate Value Theorem to the velocity function on the intervals [0. fg] and \t^. 5.5], you see that there are at
least two times during the flight when the speed was 400 miles per hour. (0 < 400 < 454.54)
112 Chapters Applications of Differentiation
49. (a) /is continuous on [— 10, 4] and changes sign,
(/(-8) > 0, /(3) < 0). By the Intermediate Value
Theorem, there exists at least one value of x in
[- 10, 4] satisfying/U) = 0.
(c)
(b) There exist real numbers a and b such that
-\0 < a < b < A and /(a) = f(b) = 2. Therefore,
by Rolle's Theorem there exists at least one number c
in (- 10, 4) such that /'(c) = 0. This is called a criti-
cal number.
(d)
8--
iX
(e) No,/' did not have to be continuous on [—10, 4].
51. /is continuous on [-5, 5] and does not satisfy
the conditions of the Mean Value Theorem.
=>/is not differentiable on (-5, 5).
Example: f{x) = |;c|
M*-i
53. False. /U) = l/j:has a discontinuity zlx = 0.
55. True. A polynomial is continuous and differentiable everywhere.
57. Suppose that p{x) = x^+ 1 + ax + b has two real roots Xj and X2. Then by Rolle's Theorem, since p{xy) = p{x-^ = 0, there
exists c in (x,, x^ such that p '(c) = 0. But p '{x) = {In + l)x^ + a t^ 0, since n > 0, a > 0. Therefore, p(x) cannot have two
real roots.
59. \fp{x) = Ax- + Bx + C, then
fib) - f{a) _ (Ab- + Bb + C) - (Aa^ + Ba + C)
p '(x) = 2Ax + B
b — a b — a
A{b^ - a2) + B{b - a)
b- a
{b - a)[A(b + a) + B]
b- a
= A(b + a) + B.
Thus, 2Ax = A{b + a) and x = (b + a)/2 which is the midpoint of [a, b].
61. fix) = 2 cos X differentiable on (— oo, oo).
f'(x) = — 2 sin X
-5 ^ /'(x) < 2 =>f'(x) < 1 for all real numbers.
Thus, from Exercise 60, /has, at most, one fixed point, (x == 0.4502)
Section 3.3 Increasing and Decreasing Functions and the First Derivative Test 113
Section 3.3 Increasing and Decreasing Functions and the First Derivative Test
1. fix) = X-- 6x + i
Increasing on: (3, oo)
Decreasing on: (—00, 3)
3. y =
}?
'ix
Increasing on: ( - 00, - 2), (2, 00)
Decreasing on: (-2,2)
5- fix) = ^
fix)
x'
Discontinuity: x = 0
Test intervals:
-00 < JT < 0
0 < .t < 00
Sign of fix):
/'>o
/'< 0
Conclusion:
Increasing
Decreasing
Increasing on (- 00, 0)
Decreasing on (0, 00)
7. six)
2x- S
g'ix) = Ix - 2
Critical number: x = 1
Test intervals:
-00 < j: < 1
1 < .r < cso
Signof g'W:
g' <Q
g' >o
Conclusion:
Decreasing
Increasing
Increasing on: (l,oo)
Decreasing on: (— 00, 1)
9. .v =
y' =
xVl6 — x^
-2(;c2 - 8)
Domain: [-4, 4]
-2
V16 - x"- yi6
Critical numbers: x = ±2^2
=ix - 2V2)(.v + 272)
Test intervals:
-4 < X < -2V2
-2v^ < .t < 2v^
2V2 < X < 4
Sign of _v ':
>-' < 0
v' > 0
y' < 0
Conclusion;
Decreasing
Increasing
Decreasing
Increasing on ( - 2 VI, 2 V2)
Decreasing on (-4, -2V5), (2^2, 4)
11- fix) = x^
6x
fix) = 2x - 6 = 0
Critical number: x = 3
13. fix) = -2x^ + 4x + ?,
fix) = -4.V + 4 = 0
Critical number: .v = 1
Test intervals:
-00 < j: < 3
3 < j: < 00
Sign of/'U):
/'< 0
/'>o
Conclusion:
Decreasing
Increasing
Increasing on: (3, 00)
Decreasing on: (-00, 3)
Relative minimum: (3, - 9)
Test intervals:
-00 < x < 1
1 < .V < 00
Sign of/'t.v)-
/'> 0
/' < 0
Conclusion:
Increasing
Decreasing
Increasing on: {-00,1)
Decreasing on: (1, 00)
Relative maximum: (1.5)
114 Chapters Applications of Differentiation
15. fix) = 2x3 + 3x2 - 12a:
fix) = 6x2 + 6x - 12 = 6(;^ + 2)(x - 1) = 0
Critical numbers: x = —2, 1
Test intervals:
-oo < X < -2
-2 < X < 1
1 < X < oo
Sign of/'(x):
/'>o
/' < 0
/'>0
Conclusion:
Increasing
Decreasing
Increasing
Increasing on: (— oo, —2), (1, oo)
Decreasing on: (-2,1)
Relative maximum: (—2,20)
Relative minimum: (1,-7)
17. fix) = x2(3 - x)
3x2
fix) = 6x — 3x2 = 3^('2 _ J,)
Critical numbers: x = 0, 2
Test intervals:
— oo < X < 0
0 < X < 2
2 < X < oo
Sign of /'(x):
/'<o
/'>0
/'<o
Conclusion:
Decreasing
Increasing
Decreasing
Increasing on: (0, 2)
Decreasing on: (-oo, 0), (2, oo)
Relative maximum: (2, 4)
Relative minimum: (0, 0)
19. /(^) =
x^ - 5x
fix) = x^ - 1
Critical numbers: x = —1,1
Test intervals:
-oo < X < -1
- 1 < X < 1
1 < X < oo
Sign of/'(x):
/'>0
f <Q
f> 0
Conclusion:
Increasing
Decreasing
Increasing
Increasing on: (— oo, — 1), (1, oo)
Decreasing on: (-1,1)
Relative maximum: (—1,5)
Relative minimum: (l, -5)
Section 3.3 Increasing and Decreasing Functions and the First Derivative Test 115
21. fix) = .r'/3 + 1
fix) = -x'^/^ = — ^
Critical number: x = 0
Test intervals:
-oo < j: < 0
0 < j: < oo
Sign of/'(-t):
r > 0
/'>0
Conclusion:
Increasing
Increasing
Increasing on: (-00,00)
No relative extrema
23. /W = (x - 1)2/3
2
/'W
3U - 1)'/'
Critical number: x = 1
Test intervals:
-co < X < 1
1 < .V < CO
Sign of/'U):
/'< 0
/'>0
Conclusion:
Decreasing
Increasing
Increasing on: (l,oo)
Decreasing on: (-00, 1)
Relative minimum: (1,0)
25. fix) = 5 - |x - 51
fix) =
x
|x- 5
Critical number: x = 5
l-l, X >
Test intervals:
-00 < X < 5
5 < jc < 00
Sign off'ix):
f'>0
/'<o
Conclusion:
Increasing
Decreasing
Increasing on: (—00, 5)
Decreasing on: (5, 00)
Relative maximum: (5. 5)
27
fix) =x + ^
fix) = 1 - A =
X-- 1
X-
Critical numbers:
x= -1,
1
Discontinuity: x
= 0
Test intervals:
- 00 < v < - 1
- 1 < .r < 0
0 < .t < 1
1 < JT < oo
Sign off'ix):
/'>0
/'<0
/'<o
/'> 0
Conclusion:
Increasing
Decreasing
Decreasing
Increasing
Increasing on: (-00, - 1). (1, 00)
Decreasing on: (-1, 0), (0, 1)
Relative maximum: (- 1, -2)
Relative minimum: (1,2)
116 Chapter 3 Applications of Dijferentiation
29. f{x)
fix)
{x^ - 9)(2.r) - (x'){2x) _ -\8x
ix^- - 9r
Critical number: x = 0
Discontinuities: x = —3, 3
(x^ - 9)2
Test intervals:
-oo < jc < -3
-3 < JC < 0
0 < X < 3
3 < X < oo
Sign of/'U):
/' > 0
/' > 0
/' < 0
/'<0
Conclusion:
Increasing
Increasing
Decreasing
Decreasing
Increasing on: (-oo, -3), (-3, 0)
Decreasing on: (0, 3), (3, oo)
Relative maximum: (0, 0)
31. fix) =
fix) =
X' - 2a: + 1
X + \
(x + l)(2x - 2) - (x^ -2x+ 1)(1) _ x^ + 2jc - 3 (x + 3)(x - 1)
ix + ir
Critical numbers: x = —3, 1
Discontinuity: ;<: = — 1
ix + IT'
ix + ir
Test intervals:
-oo < X < -3
-3 < X < -1
- 1 < .1 < 1
1 < ;c < oo
Sign of/'U):
/'> 0
/'< 0
/'<o
/'> 0
Conclusion:
Increasing
Decreasing
Decreasing
Increasing
Increasing on: (— oo, -3), (1, oo)
Decreasing on: (—3, — 1), (— 1, 1)
Relative maximum: (-3,-8)
Relative minimum: (1,0)
33. fix) = - + cosjc, 0 < X < Itt
fix) = - - sin ;c = 0
Critical numbers: x = —,—r
0 0
Test intervals:
0<x<f
0
77 577
Sir
-^ < X < 277
0
Sign of fix):
f >o
/'<o
/'> 0
Conclusion:
Increasing
Decreasing
Increasing
Increasing on: ( 0, — j, ( — , 277
Decreasing on:
77 577
6'T
Relative maximum:
Relative minimum:
77 77 + 6v3
6' 12
577 577 — 6V3
T' 12
Section 3.3 Increasing and Decreasing Functions and the First Derivative Test 117
35. fix) = sin^ X + sin X, 0 < X < Itt
fix) = 2 sin j: cos x + cos x = cos x(2 sin a: + 1 )
_ . . , , IT 7tt 37T IItt
Cntical numbers: x = —, — -, -r-. — r—
2 6 2 o
Test intervals:
0<.<f
TT 777
2<^<T
777 3t7
377 1177
2 < •'^ < 6
1177
•— - < .r < 277
6
Sign of/'(:t):
/'>0
/'< 0
/'>0
/'<0
/'> 0
Conclusion:
Increasing
Decreasing
Increasing
Decreasing
Increasing
I ^ '''■\ /777 37r\ (IItt ^
Increasing on: ( 0, -I, I — , — I, I — , 277
Decreasing on:
Relative minima:
77 7t7\ /377 II77
2' 6 y V 2
777 1\ /11t7 1
6 ' 4
Relative maxima: I — , 2 I, ( — , 0
37. f(x) = 2xV9 - x\ [-3, 3]
2(9 - 2x^-)
, 2(9 - 2^2)
Critical numbers: x = +— ^ = +
J2 2
(d) Intervals:
-3,
3^2
fix) < 0
Decreasing
3v^ 372
" 2 ' 2
/'(.x) > 0
Increasing
/'(.r) < 0
Decreasina
/is increasing when/' is positive and decreasing
when/' is negative.
39. /(r) = t^ sin f, [0, 277]
(a) fit) = t^cost + 2t sin t
= t(t cos t + 2 sin f)
(b)
(c) t(t cos / + 2 sin t) = 0
t = OoT t = - 2 tan r
f cot r = - 2
t = 2.2889. 5.0870 (graphing utility)
Critical numbers: r = 2.2889. t = 5.0870
(d) Intervals:
(0.2.2889) (2.2889,5.0870) (5.0870.277)
/'(f) > 0 /'(f) < 0 /'(f) > 0
Increasing Decreasing Increasing
/is increasing when/' is positive and decreasing when
/' is negative.
118 Chapters Applications of Differentiation
41. fix) = 5 ; = -^^ :r^ = x^ -7,x, x + ±1
fix) = gix) =x^ - 3xf0Tal\xit±i_
fix) = 3a:2 - 3 = 3(JC2 - 1), X 7t ±1 /'(j;) 7t 0
/symmetric about origin
zeros of/: (0, 0), (± 73, o)
No relative extrema
Holes at (-1,2) and (1,-2)
43. fix) = c is constant =*/'W = 0
45. /is quadratic =>/' is a line.
I I I I I
-2--
-4--
I I I I I > J
47. /has positive, but decreasing slope
I iTT
-2-
-4-
In Exercises 49-53,/'W > 0 on (-oo, -4),f'(x) < 0 on (-4, 6) and/'W > 0 on (6, oo).
51. gix) = -fix) S3, gix) = fix- W)
49. gix) = fix) + 5
8'ix)=f'ix)
g'(0)=/'(0) < 0
g'ix) = -fix)
g'i-6) = -f'i-6) < 0
g'ix) = fix- 10)
g'(0)=/'(-10) > 0
I> 0, x < 4 =>/ is increasing on (-00, 4).
undefined, x = 4
< 0, X > 4 =»/is decreasing on (4, oo).
Two possibilities for fix) are given below.
(a)
(b)
Section 3.3 Increasing and Decreasing Functions and the First Derivative Test 119
57. The critical numbers are in intervals (-0.50, -0.25) and
(0.25, 0.50) since the sign of/' changes in these intervals,
/is decreasing on approximately (-1, -0.40), (0.48, 1),
and increasing on (-0.40, 0.48).
Relative minimum when x ~ ~ 0.40.
Relative maximum when x = 0.48.
59. fix) = X, g{x) = sin x. 0 < .v < tt
(a)
fix) seems greater than gix) on (0, tt).
(b)
x > sin j: on (0, it)
X
0.5
1
1.5
T
2.5
3
fix)
0.5
1
1.5
2
2.5
3
six)
0.479
0.841
0.997
0.909
0.598
0.141
(c) Let hix) = fix) — gix) = X - sin x
h 'ix) = 1 — cos ;c > 0 on (0, tt).
Therefore, hix) is increasing on (0, it). Since
hiQ) = 0, hix) > 0 on (0. tt). Thus.
jr — sin^: > 0
X > sin.r
fi.x) > gi.x) on (0, tt).
61. V = kiR - r)r^ = kiRr^ - r^)
v' = kilRr- 3r=)
= krilR - 3r) = 0
r = 0 or f /?
Maxmium when r = f /?.
v/? R
63. P = 7— — ' p ..,, V and /?, are constant
(a, + Kj)"
rfP _ (/?, + R.JHvR,) - vR,R.[2iR, + j?,)(l)]
d/fj (^, + /?,)*'
^ v/;,(/?, - ^,) ^ ^
Maximum when ^j = /?; .
65. (a) S = 0.1198t^ - 4.4879r^ + 56.9909F - 223.0222r + 579.9541
(b) 1500
(c) S' = 0 for f = 2.78. or 1983, (311.1 thousand bankruptcies)
Actual minimum; 1984 (344.3 thousand bankruptcies)
67. (a) Use a cubic polynomial
fix) = fljx' + a2.x- + a,.v + a^.
(b) fix) = 3aj.x'- + la^x + a,
(0, 0): 0 = ao (/(O) = 0)
0 = a, (/'(O) = 0)
(2,2): 2 = 8a3 + 4fl, (/(2) = 2)
0 = 12a3 + 4a, (/'(2) = 0)
(c) The solution is Op = a, = 0. O; = -. a,
3 ,
fix) = -|.v-' + |.v
(d)
\
(2. 2)
(0. 0)
\
120 Chapters Applications of Differentiation
69. (a) Use a fourth degree polynomial /W = a^x'^ + a^x^ + a,^^ + "i-* + "^o-
(b) fix) = 4a^x^ + 3^3x2 + 2^2^ + a,
(0, 0): 0 = Qo (/(O) = 0)
0 = a, (/'(O) = 0)
(4, 0): 0 = 256^4 + 640, + IGa^ (/(4) = 0)
0 = 256a^ + 48a3 + 8^2 (/'(4) = 0)
(2, 4): 4 = I6a^ + Sa, + 4^2
0 = 32a. + 12a, + 4a,
(/(2) = 4)
(/'(2) = 0)
(c) The solution is Oq = a, = 0, a, = 4, aj = — 2, 04 = —
fix) = -x" - 2^ + Ax-
id)
\
(2.4) J
A/
(0,0)
(4.0)
71. True
Let h{x) = f{x) + g{x) where /and g are increasing. Then
h'ix) = fix) + g'ix) > 0 since /'(x) > Oandg'U) > 0.
73. False
Let/(x) = x^, then/'(x) = 3x^ and/only has one
critical number. Or. let/U) = x^ + 3x + 1, then
fix) = 3(x' + 1) has no critical numbers.
75. False. For example, fix) = x^ does not have a relative extrema
at the critical number x = 0.
77. Assume that/'(x) < 0 for all x in the interval (a, b) and let x^ < x-^ be any two points in the interval. By the Mean Value
ve know there e
fix.) -fix,)
Theorem, we know there exists a number c such that x^ < c < x^, and
. /'(c)
Since/'(c) < 0 and-nr, - x, > 0, then/(x2) — /(x,) < 0, which implies that/Cxj) < fix). Thus,/is decreasing on the
interval.
79. Let/U) = i\ + x)" - ivn - 1. Then
fix) = n(l + x)"-^ - n
= «[(! +j:)"~' - 1] > Osincex > Oandn >1.
Thus./W is increasing on (0, 00). Since /(O) = 0 =^fix) > 0 on (0, 00)
i\ + x)" - nx - I > 0 => (1 + jc)" > I + nx.
Section 3.4 Concavity and the Second Derivative Test 111
Section 3.4 Concavity and the Second Derivative Test
1. y = x'- - X — 1, y" — 2
Concave upward: (-00,00)
3. fix) =
24
^ ^ - 144(4 - x'^)
x~+ 12'-' ~ (;c2 + 12)3
Concave upward: (-00, -2), (2, 00)
Concave downward: (—2,2)
5. fix) = J^, y
4(3x^ + 1)
r ■' (x2 - 1)3
Concave upward: (-00, - 1), (1, 00)
Concave downward: (-1,1)
7. fix) = 3x^ - x^
fix) = 6x- 3x2
fix) = 6 - 6x
Concave upward: (-00. 1)
Concave downward: (1, 00)
9. v = 2x- tanx, (-^.7
y' = 2 — sec-x
>'"= —2 sec-x tanx
Concave upward: ( - — , 0
Concave downward: 0,
11. fix) = x> - 6x- + \2x
fix) = 3x- - 12x + 12
fix) = 6(x - 2) = 0 when x = 2.
The concavity changes at x = 2. (2, 8) is a f)oint of
inflection.
Concave upward: (2, cc)
Concave downward: (—00, 2)
13.
fix)
1
2x2
fix) = x3 - 4x
fix) = 3x2-4
fix) = 3x2 _ 4 = Owhenx
2
^73-
Test interval:
2
CO < X < r-
J3
2 "^
~~r^ < X < -7=
J3 V3
2
^ < X < 00
73
Sign of/"(x):
fix) > 0
/"(-v) < 0
fix) > 0
Conclusion:
Concave upward
Concave downward
Concave upward
Points of inflection:
20
'■V3'
122 Chapter 3 Applications of Differentiation
15. /(;c) = x(x - 4)3
fXx) = x[3{x - 4)2] + U - 4)3
= U - 4)^(4x - 4)
f"(x) = 4{x - \)[2{x - 4)] + 4(x - 4)2
= 4(a: - 4)[2U - 1) + (x - 4)]
= 4(x - 4)(3x - 6) = \2{x - 4){x - 2)
f"{x) = 12(jc - 4)(jc - 2) = 0 when j; = 2, 4.
Test interval:
-oo < X <1
2 < j: < 4
4 < jt < oo
Sign of /"W:
fix) > 0
fix) < 0
/"(x) > 0
Conclusion:
Concave upward
Concave downward
Concave upward
Points of inflection: (2, - 16), (4, 0)
17. f(x) = xjx + 3, Domain: [-3, oo)
/'W=.(tj(-.3)-./2.vm^|^
/"W
6v<m - 3U- + l){x + 3)-'/2 _ 3(x + 4)
4U + 3) 4{x + 3)3/2
/"(jc) > 0 on the entire domain of/ (except for x = - 3, for which/"(x) is undefined). There are no points of inflection.
Concave upward on (-3, oo)
19. /W = -
f\x) =
/"W =
;c2+ 1
1 -;c2
(;c2 + 1)2
2x(x2 - 3)
(x2 + 1)3
= 0 when a: = 0, ±V3
Test intervals:
— oo < X < — V3
-73 <;c < 0
0 < ;c < 73
73 < X < oo
Sign of/'W:
/"<o
/">0
/"<0
/">0
Conclusion:
Concave downward
Concave upward
Concave downward
Concave upward
Points of inflection:
■73,-^1 (0,0), (73,^
21. fix) = sin( - 1, 0 < a: < 47r
fix) = icos(f
/"W = -^sini 2
/"(;c) = 0 when x = Q, Itt, 4t7.
Point of inflection: (27r, 0)
Test interval:
0 < .^r < 277
277 < j: < 477
Sign of/"W:
/" < 0
/">0
Conclusion:
Concave downward
Concave upward
Section 3.4 Concavity and the Second Derivative Test 123
23. f(x) = sec(;<: - -\,Q < x < Att
fix) = secfjr - -j tanix - -
f"(x) = sec^l X - —\ + seel x - —\ tan^j x - —\ i= Q for any x in the domain of/.
Concave upward: (0, tt), {Itt, Ztt)
Concave downward: (tt, 2tt), (Stt, 47r)
No points of inflection
25. f{x) = 2 sin;<: + sin 2;c, 0 < j: < 277
fix) = 2 cos j: + 2 cos 2x
f"{x) = — 2 sin j: - 4 sin 2x = — 2 sin ;c(l +4 cos x)
f"{x) = 0 when x = 0, 1 .823, tt, 4.460.
Test interval:
0 < A- < 1.823
1.823 < .X < TT
77 < ,x: < 4.460
4.460 < .V < 277
Sign of /"(a):
/"< 0
/"> 0
/"< 0
/">0
Conclusion:
Concave downward
Concave upward
Concave downward
Concave upward
Points of inflection: (1.823, 1.452), {77, 0), (4.46, - 1.452)
27. /(a) = A^ - 4x3 + 2
/'(a) = 4a3 - 12a' = Ax-{x - 3)
/"(a) = 12a' - 24a = 12a(a - 2)
Critical numbers: a = 0, a = 3
However, /"(O) = 0, so we must use the First Derivative
Test. /'(a) < 0 on the intervals (-00, 0) and (0, 3); hence,
(0, 2) is not an extremum./"(3) > 0 so (3, -25) is a
relative minimum.
29. fix) = (a - 5)2
/'(a) = 2(a-5)
f"{x) = 2
Critical number: a = 5
/"(5) > 0
Therefore, (5, 0) is a relative minimum.
31. /(a) = a3 - 3a2 + 3
fix) = 3a2 - 6a = 3a(a - 2)
/"(a) = 6a - 6 = 6(a - 1)
Critical numbers: a = 0, a = 2
/"(O) = -6 < 0
Therefore, (0, 3) is a relative maximum.
/"(2) = 6 > 0
Therefore, (2, — 1) is a relative minimum.
33. g{x)=x\6-xy
g\x) = x{x - 6)2(12 - 5.r)
g"(x) = 4(6 - x)(5x~ - 24a + 18)
Critical numbers: x = 0, -f , 6
^"(0) = 432 > 0
Therefore, (0, 0) is a relative minimum.
g"{fj = - 155.52 < 0
Therefore, \-f, 268.7) is a relative minimum.
g"(6) = 0
Test fails by the First Derivative Test. (6. 0) is not
an extremum.
124 Chapters Applications of Differentiation
35. f(x) = .x2/3 - 3
2
fix) =
fix) =
-2
Critical number: a; = 0
However, /"(O) is undefined, so we must use the First
Derivative Test. Since /'(x) < 0 on (— oo, 0) and
f'(x) > 0 on (0, oo), (0, -3) is a relative minimum.
37. f(x) =x + -
■' X
fix) = 1 -
Critical numbers: x = ±2
f'i-2) < 0
Therefore, (—2, —4) is a relative maximum.
/"(2) > 0
Therefore, (2, 4) is a relative minimum.
39. fix) = cosx - X, 0 < x < 417
fix) = - sin ;c - 1 < 0
Therefore, / is non-increasing and there are no relative extrema.
41. fix) = O.lx^ix - 3)3, [- 1, 4]
(a) fix) = 0.2;c(5;c - 6)U - 3)^
/"W = U - 3)(4;c2 - 9.6a: + 3.6)
= OAix - 3)(10.x2 _ 24.x + 9)
(b) /"(O) < 0 =» (0, 0) is a relative maximum.
/"(f) > 0 =*• (1.2, - 1.6796) is a relative minimum.
Points of inflection:
(3, 0), (0.4652, -0.7049), (1.9348, -0.9049)
/is increasing when/' > 0 and decreasing when
/' < O./is concave upward when/" > 0 and con-
cave downward when/" < 0.
43. fix) = sin jc - -sin 3x + -sin 5x, [0, n]
(a) fix) = cos X - cos 3x + cos 5x
fix) = 0 when x = T- ^ = T' ^ = "g"-
fix) = — sin X + 3 sin 3a: — 5 sin 5x
fix) = 0 when x = ^, x = ^, x = 1.1731. .t == 1.9685
6 6
(b) /"( — 1 < 0 => ( — , 1 .53333 1 is a relative maximum.
Points of inflection: I -, 0.2667], (1.1731, 0.9638),
(1.9685,0.9637), [y"' 0-2667
Note: (0, 0) and (ir, 0) are not points of inflection
since they are endpoints.
(c)
The graph of/ is increasing when/' > 0 and decreas-
ing when/' < O./is concave upward when/" > 0
and concave downward when/" < 0.
Section 3.4 Concavity and the Second Derivative Test 125
45. (a)
/' < 0 means/decreasing
/' increasing means
concave upward
(b)
/' > 0 means/ increasing
/' increasing means
concave upward
47. Let/U) = ;d.
f"(x) = Ux^
/"(O) = 0, but (0, 0) is not a point of inflection.
49.
51.
53. >
55. y
57.
/"is linear.
/' is quadratic.
/is cubic.
/concave upwards on (— oo. 3), downward on (3. oo).
126 Chapters Applications of Differentiation
59. (a) n = 1:
fix) = 1
fix) = 0
No inflection points
n = 2:
/W = (;c - 2)2
/'W = 2U - 2)
/"W = 2
No inflection points
Relative minimum:
(2,0)
n = 3:
fix) = (x- ly
fix) = 3{x - If
fix) = 6{x - 2)
Inflection point: (2, 0)
,/
Point of
inflection
n = 4:
/W = {x- ly
fix) = 4(;c - 2)3
/W = 12(;c - 2)2
No inflection points:
Relative minimum:
(2,0)
J.
Conclusion: If n > 3 and n is odd, then (2, 0) is an inflection point. If n > 2 and n is even, then (2, 0) is a relative minimum.
(b) Let fix) = ix- 2)", fix) = nix - 2)"~\ fix) = n(« - \)ix - 2)"-\
For n > 3 and odd, « — 2 is also odd and the concavity changes at x = 2.
For n > 4 and even, n — 2 is also even and the concavity does not change at j: = 2.
Thus, j: = 2 is an inflection point if and only if « > 3 is odd.
61. fix) = ax^ + bx^ + ex + d
Relative maximum: (3, 3)
Relative minimum: (5, 1)
Point of inflection: (4, 2)
fix) = 3ax^ + 2bx + cf'ix) = 6ax + 2b
/(3) = 27. + % + 3c + rf=3 ,„._,..^..__._49^^8,^^^_1
/(5) = 125a + 25b + 5c + d= 1
/'(3) = 27a + 6fc + c = 0, /"(4) = 24a + 2fc = 0
- 49a + 8fc + c = - 1 24a + 2* = 0
27a + 6fc + c = 0 22a + 2fc = - 1
22a + 2b = - 1 2a
a = ii)= -6,c = f,rf= -24
fix) = \x^ - 6x^ + f ;c - 24
= 1
Section 3.4 Concavity and the Second Derivative Test 127
63. f(x) = ax^ + bx^ + ex + d
Maximum: (-4, 1)
Minimum: (0, 0)
(a) f'(x) = 3ax- + 2bx + c, f"{x) = 6ax + 2b
m = 0=^d = 0
/(-4) = 1 =
/'(-4) = 0 =
/'(0) = 0 =
Solving this system yields a = j^ 3"^ i> = 6a = ^.
fix) = j,x^ + ^,x^-
-64a + 16fc - 4c = 1
48a - 8b + c = 0
c = 0
(b) The plane would be descending at the greatest rate at
the point of inflection.
fix) = 6ax + 2b = j^x + ^
Two miles from touchdown.
0
x = -2.
65. D = 2x' - 5Lx^ + 3L-x^
D' = 8.r3 - \5Lx^ + 6L~x = .r(8.x- - 151.x + 6L-) = 0
15L± 733Z. /l5 ± 733 V
, = 0or.. = =(^_^^JZ.
By the Second Derivative Test, the deflection is
maximum when
15
16
L = 0.578L.
67. C = 0.5.V- + 15.r + 5000
C^^ = 0.5..-.15^^
X X
C = average cost per unit
^.0.5-^ = 0when.v=100
dx X-
By the First Derivative Test, C is minimized when
X = 100 units.
69.
sXt) =
S"(t) =
5000P
8 + ^2
80,000f
(8 + t^r
80,000(8 - it-)
(8 + a'
S"(t) = 0 for r = ^/873 « 1.633.
Sales are increasing at the greatest rate at f = 1.633 years.
71. /(;c) = 2(sin.r + cos;c),
f'(x) = 2(cos X — sin .r),
f"{x) = 2{-sin.v - cos.t),
P,(jc) = 2v^ + o(x - Jj = 2V^
/( - I = 272
0
/"(f) = -^
Pi'{x) = 0
P,(x) = 272 + o(.v - fl + ^(-272)(.v - f )^ = 2 V2 - .^(.v - ^
PAx) = -272(.r - fl
P^x) =-lJl
The values of/, P,, Z'^, and their first derivatives are equal at .r = ttJA. The values of the second derivatives of/ and P, are
equal at x = Tr/4. The approximations worsen as you move away from x = 7r/4.
128 Chapters Applications of Differentiation
73. /U) = Vl -X,
1
/'W = -
fix) = -
2Vl - X
1
4(1 - .t)3/2'
/(O) = 1
/'(o) = 4
/"(O)
p,W = i + (-|)u-o) = i-|
PAx) = -i-
^
s^. .
/.
N
/>.'(.) = 4- J • .
P.'U) = -|
The values of/, Pj, P,' ^°d their first derivatives are equal at x = 0. The values of the second derivatives of/ and P2 are equal
at j: = 0. The approximations worsen as you move away from x = 0.
75. fix) = X sin
fix)
r COSi
x^ \x
1 n\ . n
--cos - + sm -
X \xl \x
fix) = --
x
i ^'"(i
+ sin
+ -^ cos ^ cos - = — T sm -
X \xl X- \xj x' \x.
\\ 1
A
./^
V
^(J..)
Point of inflection: | — , 0
When X > I/tt,/' < 0, so the graph is concave downward.
77. Assume the zeros of/ are all real. Then express the function as/W = a(x — r^)ix — r-^ix — r^) where r^, r^, and rj are the
distinct zeros of/. From the Product Rule for a function involving three factors, we have
fix) = a[ix - r,)(x - r^ + ix - r,)(jf - r^) + (x - r^ix - r^)]
fix) = a[ix - r,) + ix-r,J + ix- r,) + (x - r^) + (x - r^) + (x - r^)]
■ = a[6x — 2(ri + rj + rj)].
Consequently, fix) = 0 if
2(''i + ''2 + '"3) '"1 + '"•' + '"3 /
jc = 7 = f = (Average of r,, r,, a«d rj).
79. True. Lety = ax^ + bx^ + ex + d. a ¥= 0. Then y" = 6(Xt + 2fe = 0 when x= - ib/3a)
and the concavity changes at this point.
Section 3.5 Limits at Infinity 129
81. False.
fix) = 3 sin Jt + 2 cos x
fix) = 3 cos j: - 2 sin X
3 cos .V — 2 sin X = 0
3 cos j: = 2 sin x
2 = tan.r
Critical number: x = tan '(2)
/(tan"' 2) ~ 3.60555 is the maximum value of .v.
Section 3.5 Limits at Infinity
83. False. Concavity is determined by /".
1. f(x) =
3x2
.t2 + 2
No vertical asymptotes
Horizontal asymptote: v = 3
Matches (f)
3. fix.
X- + 2
No vertical asymptotes
Horizontal asymptote: y = 0
Matches (d)
5. fix)
4sin j:
X- + 1
No vertical asymptotes
Horizontal asymptotes: y = 0
Matches (b)
7. f(x) =
4x + 3
/ \-^l
2x- 1
X
10°
10'
102
103
lO-*
10^
10^
fix)
7
2.26
2.025
2.0025
2.0003
2
2
lim fix) = 2
9. fix)
-6x
V4? + 5
X
10"
10'
10-
103
10"
10-^
10«
fix)
-2
-2.98
-2.9998
-3
-3
-3
-3
lim/W = -3
11. fix) = 5 - -
1
.V2+ 1
x
10"
10'
10-
103
10*
10^
10^
fix)
4.5
4.99
4.9999
4.999999
5
5
5
lim fix) = 5
130 Chapters Applications of Dijferentiation
13. (a) Mx) = — -r- = ; = 5a - 3 H — 7
x-^ x^ x^
lim h(x) — 00 (Limit does not exist)
-r— ♦00
lim h(x) = 5
f{x) ^ 5x3 - 3;c^+ 10 _ 5 3 ^ iQ
lim /lU) = 0
. ^2 + 2
Jr^ooX" — 1
^2 + 2
15. (a) lim ,
... I v^
(b) lim
J:~*o:
(c) lim
Jr'-»oo X^ — \
x^ + 2
X - 1
= 0
= 1
= 00 (Limit does not exist)
5 - 2x3^2
17. (a) lim - , , = 0
(b) lim
(c) lim
5 - 2t3/2 _ _2
3x3/2 - 4 3
5 -2x^/2
3x-4
= — 00 (Limit does not exist)
10 r ^- 1 ,• 2 - (1/x) 2-0 2
19. lim = lim -^ , . = r r = r
i-^00 3x + 2 x->c^ 3 + (2/x) 3+0 3
2L lim
X — >oD X~ ~ 1
lim
lA _o ^
1 - (1A2) 1
23. lim
5x2
lim
5x
x + 3 x'-X-a= 1 + (3/x)
Limit does not exist.
25. lim —r^=
j:->-oo ^x'^ — X
lim ^
( forx < 0 we have x = — J]?)
= lim
-1
x^-x^i -d/x)
= -1
2x + 1
27. Iim — , .. = lim
2 +
*-=<= Vx2 — X Jr->-oo / .Jx^
(for X < 0, X = - 7?)
lim — , ,= = — 2
^^°o Vx + O/x)
29. Since (- 1/x) < (sin(2x))/x < (1/x) for all x =5^ 0, we
have by the Squeeze Theorem,
,. 1 ,. sin(2x) ^ ,. 1
lim — < lim < lim -
X -*co X X -*oo X X -*oo X
0 < lim ^^^^ < 0.
Jr-»oo X
sin(2x)
Therefore, lim = 0.
X -*oc X
3L lim
1
X ->oo 2x + sin X
Section 3.5 Limits at Infinity 131
33. (a) fix)
x+ 1
lim — V—
x->aoX + 1
lim
= -1
>x+ 1
Therefore, y = 1 and y =
1 are both horizontal asymptotes.
.^M
,, ,. .1 ,. smt
35. lim jr sm — = lim = 1
x^ac X (-»0* t
(Let a: = 1/r.)
37. lim (a: + V;c= + s) = lim
= lim
^-»-°=a: - Va:2 + 3
39. lim {x - Vat + x)
lim
- v? +
+ vVT:
= lim
X + Vx^ + X
-1
t ^°° a: + V-t- + X X-.00 1 + Vl + (lA)
41.
JC
10"
10'
102
103
10^
10^
10«
/w
1
0.513
0.501
0.500
0.500
0.500
0.500
lim (x — ^x(x - 1)) = lim
— Vj^ — X X + V-t- — x
= lim
x+ v^
^=»x + Vx^ - X
1
'™ — / 7—r
■< ^=° 1 + Vl - (1/x
2
^
43.
X
10«
10'
10-
10-'
10^
10^
10*
/(x)
0.479
0.500
0.500
0.500
0.500
0.500
0.500
Letx = 1/f.
lim X sin r-
,. sin(f/2) 1 sin(f/2) 1
= lim = hm — 7- — = -
, -,0* f r ^0* 2 f/2 2
ri^/::
132 Chapters Applications of Differentiation
45. (a)
(b) lim fix) = 3 lim f'(x) = 0
(c) Since lim f{x) = 3, the graph approaches that of a
horizontal line, lim f'{x) = 0.
47. Yes. For example, let/(x) =
6U - 21
j(x - 2)2 + r
49. >- =
2 +x
1 -;c
Intercepts: (-2,0), (0,2)
Symmetry: none
Horizontal asymptote: y = — 1 since
lim
2+ X
1 = lim .
X ->oo \ — X
»-oo 1 — X
Discontinuity: x = 1 (Vertical asymptote)
x^ - 4
Intercept: (0,0)
Symmetry: origin
Horizontal asymptote: y = 0
Vertical asymptote: x = +2
53. y
x^ + 9
Intercept: (0,0)
Symmetry: y-axis
Horizontal asymptote: y = 1 since
lim
= 1 = lim
Relative minimum: (0, 0)
4-
3--
2
1
1 2 3
Section 3.5 Limits at Infinity
133
55. y =
2x2
Intercept: (0,0)
Symmetry: y-axis
Horizontal asymptote: y = 2
Vertical asymptote: x = ±2
57. xy- = 4
Domain: a: > 0
Intercepts: none
Symmetry: x-axis
Horizontal asymptote: y = 0 since
lim -^ = 0
lim —
Vx-
Discontinuity: x = 0 (Vertical asymptote)
59. V =
2x
1 - X
Intercept: (0, 0)
Symmetry: none
Horizontal asymptote: y = —2 since
1- 2r . ,. 2x
urn = — 2 = lim .
j:-»-oo I — X .r->oo i — X
Discontinuity: x = 1 (Vertical asymptote)
H — I — I — l-»-i
2 3 4 5
61. V
Intercepts: (±v'^. O)
Symmetry: y-axis
Horizontal asymptote: y = 2 since
lim (2 - 4j ) = 2 = lim (2 - V\-
x->-ooV X-/ x^oo\ X-j
Discontinuity: .r = 0 (Vertical asymptote)
63. V = 3 + -
X
2 2
Intercept: v = 0 = 3 H — ^- =
XX
Symmetry: none
Horizontal asymptote: y = 3
Vertical asymptote: x = 0
.,.-|-|o
134 Chapters Applications of Differentiation
65. y =
Domain: (—00, —2), (2, 00)
Intercepts: none
Symmetry: origin
Horizontal asymptote: none
Vertical asymptotes: x = ±2 (discontinuities)
I -°
I 16
1 12
11/
I I I
-5-4-3-2-1
A
-12
-16--
-20--
67. /W = 5 - A = ^^^
X- x'-
Domain: (— 00, 0), (0, 00)
f'(x) = 3 => No relative extrema
/"W
''"'"^ = ---J =i. No points of inflection
Vertical asymptote: ;c = 0
Horizontal asymptote: y = S
|y = 5|
V = o|
6
\
r~
69. f{x) =
/'W =
/"U) =
x^ - ^
U^ - 4) - ;t(2jc)
{x^ - 4)2
-U2-H4)
=i^ 0 for any x in the domain of/.
+ (;c2 + 4]
(x^ - 4)2
{x^ - 4)2
(;c2 - 4)2(-2;c) + (x^ + 4)(2)(x2 - 4)(2;c)
lx{x^ + 12)
= 0 when ;t = 0.
L
r^-'?!-^ ^"
(a:2 - 4)3
Since/"(;c) > 0 on (-2, 0) and/"(;«:) < 0 on (0, 2), then (0, 0) is a point of inflection.
Vertical asymptotes: x = ±2
Horizontal asymptote: y = Q
71. /U)
.X - 2
x-2
;c2 - 4;c + 3 U - 1)(a: - 3)
^ (x2 - 4x + 3) - (;c - 2)(2;c - 4) ^ -x^ -h 4j: - 5
■^ ^""^ (x2 - 4x + 3)2 (x2 - 4x + 3)2 "^ "
/"W
(x2 - 4jc -t- 3)2(-2x + 4) - (-^2 + 4x - 5)(2)(j:2 - 4x + 3)(2x - 4)
i^^
"^
^Mi^
2(^3 - 6^2 -F 15x - 14)
(jc2 - Ax + 3)*
= 0 when x = 1.
(x2 - 4x + 3)3
Since/"U) > 0 on (1, 2) and/"U) < 0 on (2, 3), then (2, 0) is a point of inflection.
Vertical asymptote: x = l,x = 3
Horizontal asymptote: >> = 0
Section 3.5 Limits at Infinity 135
73. /W =
fix) =
fix) =
3;c
jAx'^ + 1
3
(4x2 + 1)3/2
-36x
=> No relative extrema
(4^2 + 1)V2
Point of inflection: (0,0)
Horizontal asymptotes: y —
No vertical asymptotes
0 when jc = 0.
'^ I
nS
75. g(x) = sin
X- 2
3 < X < oo
-2 cos
g\x)
X- 2
(^ - 2?
Horizontal asymptote: y = \
„ , . . X 77
Relative maximum: = —
X - 2 2
No vertical asymptotes
277
77- 2
5.5039
77. /(x)
(a)
x^ - 3x- + 2
x(x — 3)
^(x) = X +
x(x - 3)
■M
/=
(b) /(x) =
t^ - 3x- + 2
x(x - 3)
xKx - 3) ^ 2
x(x — 3) x(x — 3)
9
= X +
x(x - 3)
g{x)
(c)
fT?T-')
/
/ r^^ 1
■J 1 V = sm( 1 ) 1
■ — ' — ' — ■ — ' — ■ — ■ — ' — ' — ■
The graph appears as the slant asymptote v = x.
79. C = 0.5x + 500
C = ^
X
C = 0.5 +
lim 10.5 +
500
X
500
= 0.5
136 Chapters Applications of Differentiation
81. line: inx - y + 4 - 0
12 3 4
(a) d =
|Ati + B,vi + C\ _ \m{3) - 1(1) + 4|
|3m + 3 1
Vm2 + 1
V"r + 1
(b)
(c) lim (f(m) = 3 = lim dim)
The line approaches the vertical Une x = 0. Hence, the
distance approaches 3.
83. (a) TiU) = -0.003/2 + o.677; + 26.564
(b)
(c)
r, =
1451 + 86r
58 + /
(d) r,(0) = 26.6
TjCO) » 25.0
86
(e) lim r.
1
86
(f) The limiting temperature is 86.
r, has no horizontal asymptote.
85. Answers wiU vary. See page 195.
87. False. Let/U) =
Vx^ + 2
(See Exercise 2.)
Section 3.6 A Summary of Curve Sketching
1. /has constant negative slof>e. Matches (D)
3. The slope is periodic, and zero at j: = 0. Matches (A)
5. (a) fix) = OfoTx = -2 and^; = 2
/' is negative for — 2 < ,t < 2 (decreasing function).
/' is positive for x > 2 and x < — 2
(increasing function).
n?) fix) = 0 at X = 0 (Inflection point).
/"is positive for x > 0 (Concave upwards).
/" is negative for X < 0 (Concave downward).
(c) /' is increasing on (0, oo). (/" > 0)
(d) /'(x) is minimum at x = 0. The rate of change of/ at
X = 0 is less than the rate of change of /for all other
values of x.
Section 3.6 A Summary of Curve Sketching 137
7. y =
y =
X- + 3
6x
{x^ + 3)2
18(1 - .Q
= 0 when x = 0.
= 0 when x = ± 1 .
Horizontal asymptote: y = 1
i—h-x
V
v'
y"
Conclusion
-oo < X < -\
-
-
Decreasing, concave down
x= -I
1
4
-
0
Point of inflection
-1 < X < 0
-
+
Decreasing, concave up
x = 0
0
0
+
Relative minimum
0 < X < \
+
+
Increasing, concave up
x= 1
1
4
+
0
Point of inflection
1 < X < oo
+
-
Increasing, concave down
9. y =
- 3
1
:j < 0 when x ^ 2.
U - 2)
^ U - 2)3
No relative extrema, no points of inflection
Intercepts: (-, o), (o,-y
Vertical asymptote: x = 1
Horizontal asymptote: v = — 3
11. y =
Ix
< Oifxifc ±1.
X- - 1
, -2ix- + 1)
„ 4x(x- + 3)
Inflection point: (0, 0)
Intercept: (0, 0)
Vertical asymptote: a = ± 1
Horizontal asymptote: y = 0
Symmetry with respect to the origin
138 Chapters Applications of Differentiation
13. g(;c)=., + -
g'(x) = 1 -
.V2+ 1
{x^ + 1)2
{X^ + 1)2
V3
Owhenx = 0.1292, 1.6085
„, , 8(3x2 _ 1)
^ W = (^ + 1)3 = 0 when X = ± 2
g"(0.1292) < 0, therefore, (0.1292, 4.064) is relative maximum.
^"(1.6085) > 0, therefore, (1.6085, 2.724) is a relative minimum.
Points of inflection:
2 ,2.423j,(^^, 3.577
Intercepts: (0,4), (- 1.3788, 0)
Slant asymptote: y = x
(0.1292. 4.064)
(-f . 2,423)
y
(1.6085. 2.724)
(-1.3788.
H — f-<4
15. /(x) = ^^ = X + -
.r X
/'(x) = 1 - -r = 0 when x = ± 1.
x^
fix) =4^0
x'
Relative maximum; (—1, —2)
Relative minimum: (1,2)
Vertical asymptote: x = 0
Slant asymptote: y = x
-4 -2 '
(1,2)
2
(-1.-2)
x2 - 6x + 12 4
17. y = : = X — 2 +
y'= 1
x-4
4
x-4
{x - 4)2
(x - 2)(x - 6)
ix - 4)2
0 when x = 2, 6.
y
ix - 4)3
j" < 0 when x = 2.
Therefore, (2, —2) is a relative maximum.
y" > 0 when x = 6.
Therefore, (6, 6) is a relative minimum.
Vertical asymptote: x = 4
Slant asymptote: y = x - 2
(0,-3) J,
A' y = ^
(6.6)
■7^ — I 1 1 h-^'
6 8 10
(2,-2)
Section 3.6 A Summary of Curve Sketching 139
19. .y = xjx - 4,
Domain; (-oo, 4]
y ' = — , = 0 when x = - and undefined when x = 4.
Bx - 16
16
J - .,. M ;-> = 0 when x = ^- and undefined when x = 4.
4(4 — x)^'^ 3
Note: X = y is not in the domain.
y
y '
y"
Conclusion
8
-oo < X < -
+
-
Increasing, concave down
8
"=3
16
3V3
0
-
Relative maximum
8
- < X < 4
-
-
Decreasing, concave down
x = 4
0
Undefined
Undefined
Endpoint
21. hix) = xV9 - x= Domain:
9 - 2x-
/! '(x) = , = 0 when x = ±
V9 - X'
3 < X < 3
3 372
72
h"{x) = 1^' ..^? = 0 when x = 0
(9 - x2)3/:
Relative maximum:
3V^ 9
Relative minimum: —
2 '2
372 9
2 ' 2;
Intercepts: (0, 0), (±3, 0)
Symmetric with respect to the origin
Point of inflection: (0, 0)
23. y = 3x2/3 _ 2x
y' = 2x-'/3 - 2 =
2(1 - x'/3)
_l/3
= 0 when x = 1 and undefined when x = 0.
-2
y =
3x^/3
< 0 when x ^ 0.
y
y'
y"
Conclusion
-oo < X < 0
-
-
Decreasing, concave down
X = 0
0
Undefined
Undefined
Relative minimum
0 < X < 1
+
-
Increasing, concave down
X = 1
1
0
-
Relative maximum
1 < X < oo
-
-
Decreasing, concave down
140 Chapters Applications of Differentiation
25. y = x^-3x- + 3
y ' = 3x- — 6x = 3x{x - 2) = 0 when x = Q, x = 1
y" = 6j: - 6 = 6(a- - 1) = 0 when x = 1
y
y'
y"
Conclusion
-CXD < .T < 0
+
-
Increasing, concave down
;c = 0
3
0
-
Relative maximum
0 < X < \
-
-
Decreasing, concave down
x= 1
1
-
0
Point of inflection
1 < ;c < 2
-
+
Decreasing, concave up
x = 2
-1
0
+
Relative minimum
1 < X < CO
+
+
Increasing, concave up
(-0.879, 0)
(2, -1)
-(1.347.0)
27. y = l- X- x^
y'= -\-3x'^
No critical numbers
y" = — 6x = 0 when x = Q.
y
y '
y"
Conclusion
-oo < X < Q
-
+
Decreasing, concave up
x = Q
2
-
0
Point of inflection
0< j: < oo
-
-
Decreasing, concave down
29. f{x) = 3.r3 - 9;c + 1
fix) = 9;c2 - 9 = 9(;c2 - i) = o whenx = ±1
f"{x) = \%x = Owhenjc = 0
fix)
fXx)
/"(-t)
Conclusion
-oo < Jf < — 1
+
-
Increasing, concave down
.X = -1
7
0
-
Relative maximum
-1 < ;t < 0
-
-
Decreasing, concave down
x = Q
1
-
0
Point of inflection
0 < X < 1
-
+
Decreasing, concave up
X = 1
-5
0
+
Relative minimum
1 < .)C < oo
+
+
Increasing, concave up
(0.112,0)
Section 3.6 A Summary of Curve Sketching 141
31. y = 3y + 4^3
y' = 12r^ + llx' = \2x-(x + 1) = 0 when.r = 0, jc = - 1.
y"= 36x2 + 24.t = 12.t(3.r + 2) = 0 whenx = 0, jr = -f.
y
y '
y"
Conclusion
- OO < .V < - 1
-
+
Decreasing, concave up
x= -1
-1
0
+
Relative minimum
-1 < X < -f
+
+
Increasing, concave up
1
X = -f
16
21
+
0
Point of inflection
-| < X < 0
+
-
Increasing, concave down
x = 0
0
0
0
Point of inflection
0 < X < OO
+
+
Increasing, concave up
33. fix) = x^ - 4x3 + 16x
fix) = 4.r3 - 12t2 + 16 = 4(x + l)(x - 2)= = 0 whenx = - l,x = 2.
fix) = \Zx- - 24x = 12r{x - 2) = 0 when x = 0, x = 2.
/(-v)
/'W
fix)
Conclusion
- OO < X < - 1
-
+
Decreasing, concave up
x= -1
-11
0
+
Relative minimum
- 1 < X < 0
+
+
Increasing, concave up
x = 0
0
+
0
Point of inflection
0 < X < 2
-
-
Increasing, concave down
X = 2
16
0
0
Point of inflection
2 < X < OO
+
+
Increasing, concave up
(-1.679. 0)
35. y = x5 - 5x
y'= 5x* - 5 = Six* - 1) =
y" = 20x3 = 0 when x = 0.
0 whenx = ±1.
y
y'
y"
Conclusion
-OO < X < -1
+
-
Increasing, concave down
x= -1
4
0
-
Relative ma.ximum
- 1 < X < 0
-
-
Decreasing, concave down
x = 0
0
-
0
Point of inflection
0 < X < 1
-
+
Decreasing, concave up
x= 1
-4
0
+
Relative minimum
1 < X < OO
+
+
Increasing, concave up
142 Chapters Applications of Differentiation
37. y = \lx - 3|
, 2(2x - 3) . . . ^ 3
y — t; TT undefined aXx = -.
^ \lx - 3| 2
)>"= 0
y
y'
Conclusion
-OO < X < 2
-
Decreasing
3
X = 2
0
Undefined
Relative minimum
5 < .V < OO
+
Increasing
39. y = sin X — — sin 'ix, Q < x < 2tt
18
1 , „ , 77 3-77
>> = COS x — - COS 3j: = 0 when x = —, -^.
0 2 i
,, . 1 . - . , ^ TT Stt Itt IItt
>> = - sm j: + - sin 3x = 0 when a: = 0, — , —-, tt, — -, — — .
2 6 6 6 6
Relative maximum
\2'18/
Relative minimum:
377 _29
2 ' 18
Inflection points: ( -^, - j, ( ^, ^ ), (tt, 0),
TT 4\ /Stt 4
6'9/'V 6'9
7tt _4\ /JJjr _4
6' 9M 6 ■ 9
_,^ _ TT TT
41. y = 2x — tan .r, — — < j: < —
^ 2 2
43. y — 2(cscjc + sect), 0 < jr <
y ' = 2 - sec- x = 0 when x = ±— .
4
y " = — 2sec^ j: tan :«: = 0 when j: = 0.
Relative maximum: \—,— — 1
\4' 2
Relative minimum: ——,1 — x
\ 4' 2
Inflection point: (0, 0)
Vertical asymptotes: x = +—
y ' = 2(sec j: tan j: — esc x cot a:) = 0 => jc = 7r/4
Relative minimum: I — , A^Jl
Vertical asymptotes: x = 0, x = —
Section 3.6 A Summary of Curve Sketching 143
__ , . 377 377
45. g{x) = x\.znx, — — < ^ < ^
„ . X + smx cos X ^ ,
g W = ; = 0 when x = 0
g"W
cos^x
2(cosa: + jr sin;c)
Vertical asymptotes: x = -
BtT it TT 377
2 ' 2' 2' 2
Intercepts: (- 77, 0), (0, 0), (77, 0)
Symmetric with respect to y-axis.
Increasing on ( 0, — j and ( — , —
Points of inflection: (±2.80, 0)
47. f(x)
2Qx 1 \9x^ - 1
;c2 + 1 X x{x- + 1)
X = 0 vertical asymptote
y = 0 horizontal asymptote
Minimum: (-1.10, -9.05)
Maximum: (1.10,9.05)
Points of inflection: (-1.84, -7.86). (1.84, 7.86)
49. y
Vjc2 + 7
2
^^^^
(0, 0) point of inflection
y = ±1 horizontal asymptotes
51. /is cubic.
/' is quadratic,
/"is linear.
S3.
(any vertical translate of/ will do)
144 Chapters Applications of Differentiation
55.
(any vertical translate of/ will do)
57. Since the slope is negative, the function is decreasing on
(2, 8), and hence/(3) > /(5).
59. fix
4{x - 1)2
jc2 - 4jt + 5
Vertical asymptote: none
Horizontal asymptote: y = 4
^
The graph crosses the horizontal asymptote >» = 4. If a
function has a vertical asymptote six = c, the graph
would not cross it since /(c) is undefined.
61. h(x) =
6 -
2x
3 -
■ X
2(3
-x)
2, if jc ^ 3
Undefined, if a: = 3
3 -X
The rational function is not reduced to lowest terms.
j;2 - 3r - 1 3
63. fix) = ^^^ = -X+1 +
x-2
x-2
\
\
\
i \
The graph appears to approach the slant asymptote
y = -.V+ 1.
hole at (3, 2)
65. fix) = . ■ ■ (0, 4)
vGc^+T'
(a)
\.
'J\/x
^^
On (0, 4) there seem to be 7 critical numbers:
0.5, 1.0, 1.5,2.0,2.5,3.0,3.5
,, > _ —cos Tr<:(xcos TTX + lirix^ + l)sin ttx) _
(t>) / W - (^2 _^ jp/2 - "
Critical numbers = ^, 0.97, |, 1.98. |, 2.98, |.
The critical numbers where maxima occur appear to
be integers in part (a), but approximating them using
/' shows that they are not integers.
Section 3.7 Optimization Problems 145
67. Vertical asymptote: x = 5
Horizontal asymptote: v = 0
1
y = z — ?
69. Vertical asymptote: x = 5
Slant asymptote: y = 3jc + 2
1 3x~ - 13x - 9
y = 3x + 2 +
x-5
X- 5
71. fix) =
(.V - bV
(a) The graph has a vertical asymptote at x = b. If
a > 0, the graph approaches oo as x-^b.\i a < 0,
the graph approaches — oo as x — > Z?. The graph
approaches its vertical asymptote faster as \a\ — >0.
(b) As b varies, the position of the vertical asymptote
changes: x = b. Also, the coordinates of the
minimum [a > 0) or maximum [a < 0) are changed.
73. fix) =
3y
.r^ + 1
(a) For n even, /is symmetric about the y-axis. For n odd,
/is symmetric about the origin.
(b) The A-axis will be the horizontal asymptote if the
degree of the numerator is less than 4. That is,
n = 0, 1,2,3.
(c) « = 4 gives y = 3 as the horizontal asymptote.
(d) There is a slant asymptote y = 3j: if n = 5:
3;c^
X* + 1
= 3.r
3.r
(e)
X* +
n
0
1
2
3
4
5
M
1
O
3
2
1
0
N
2
3
4
5
2
3
75. (a) 2750
(b) When t= 10, MlO) = 2434 bacteria.
(c) N is a maximum when t ~ 7.2 (seventh day).
(d) N%t) = 0 for f - 3.2 .
13 250
(e) lim Nit) = -^ — = 1892.86
(-►oo 7
Section 3.7 Optimization Problems
1. (a)
First Number, x
Second Number
Product, P
10
110 - 10
10(110- 10) = 1000
20
110-20
20(110 - 20) = 1800
30
110- 30
30(110 - 30) = 2400
40
1 10 - 40
40(110 - 40) = 2800
50
1 10 - 50
50(110 - 50) = 3000
60
110-60
60(110 - 60) = 3000
—CONTINUED—
146 Chapters Applications of Differentiation
1. — CONTEWED—
(b)
First Number, x
Second Number
Product, P
10
110 - 10
10(110 - 10) = 1000
20
110 - 20
20(110- 20) = 1800
30
110- 30
30(110 - 30) = 2400
40
110-40
40(110-40) = 2800
50
110- 50
50(110- 50) = 3000
60
110- 60
60(110 - 60) = 3000
70
110- 70
70(110 - 70) = 2800
80
1 10 - 80
80(110 - 80) = 2400
90
1 10 - 90
90(110- 90) = 1800
100
110 - 100
100(110 - 100) = 1000
The maximum is attained near jc = 50 and 60.
(c) P = 41 10 - x) = llOx - ;c^
(d) MOO
(e) ^ = 1 10 - 2;c = 0 when x = 55.
ax
dx'-
= -2 < 0
The solution appears to be x = 55.
P is a maximum when x = 1 10 — jf = 55.
The two numbers are 55 and 55.
3. Let X and y be two positive numbers such that xy = 192.
S = x+y = x-\
■ X
dx
d^S 384
192
= Owhen.1 = V192.
> 0 when x = V192.
dx~ x^
5 is a minimum when x = y = Vl92.
5. Let X be a positive number.
S = x + -
-r- = I ? = 0 whenx = 1.
dx x'-
-pr = -J > 0 when x= \.
dx- j^
The sum is a minimum when jc = 1 and l/;t = 1.
7. Let X be the length and y the width of the rectaiigle.
lx + ly= 100
y = 50 - X
A = xy = .)c(50 — x)
dA
^- = 50 - 2a: = 0 when x = 25.
dx
d^-A
dx^
= -2 < Owhen^; = 25.
A is maximum when j: = _y = 25 meters.
9. Let x be the length and y the width of the rectangle,
xy = 64
64
P = Zx + 2y = 2x + 2(^) = 2x + ^
£/P , 128 „ ^
-— = 2 — — 0 when x = 8.
dx X-
d^P 256
> 0 when x = 8.
dx'^ x3
P is minimum when x = y = 2> feet.
Section 3.7 Optimization Problems 147
11. d = J{X - 4)2 + (v^ - 0)2
^
7x+ 16
Since d is smallest when the expression inside the radical
is smallest, you need only find the critical numbers of
fix) = 3^ -lx+ \6.
fix) = 2x-l = 0
By the First Derivative Test, the point nearest to (4, 0) is
(7/2, 7772).
y
4
3-
te^/x)
13. d = V(x - If + [x-^ - (1/2)P
= Jx* - Ax + (17/4)
Since d is smallest when the expression inside the radical
is smallest, you need only find the critical numbers of
f{x) =x^-4x + T-
f'(x) = 4;c3 - 4 = 0
X = 1
By the First Derivative Test, the point nearest to (2. ^) is
(1, 1).
3 (4,0)
15. ^ = fcr((2o -x) = kQoK - kx^
d^Q
-^ = fcgo - 2kx
= k{Qo - 2x) = 0 when ;c = %
^=-2k < Owhen.r = %
dx> 2
dQ/dx is maximum when x = Qq/2.
17. xy = 180,000 (see figure)
360.000\
S = X + 2y = \x + ■
of fence needed.
where 5 is the length
dS
= 1
360,000
dx X-
d-S 720.000
0 when x = 600.
dx-
x^
> 0 when x = 600.
5 is a minimum when .r = 600 meters and v = 300
meters.
19. (a) A = 4(area of side) + 2(areaofTop)
(a) A = 4(3)(11) + 2(3)(3) = 150 square inches
(b) A = 4(5)(5) + 2(5)(5) = 150 square inches
(c) A = 4(3.25)(6) + 2(6)(6) = 150 square inches
150 - 2x2
(c) S = Axy + Zx' = 150:
4x
,^ , 4\5Q-2x^\ 75 1 ,
V = x-y = x'\ = ^r A' ~ :t •'^
V 4.x: / 2 2
r = y-|.r2 = 0^.r = ±5
(b) V = (length)(width)(height)
(a) V = (3)(3)(I1) = 99 cubic inches
(b) V = (5)(5)(5) = 125 cubic inches
(c) V = (6)(6)(3.25) = 117 cubic inches
\^\
^
,'
y^
By the First Derivative Test, .r = 5 yields the maximum volume. Dimensions: 5 x 5 x 5. (A cube!)
148 Chapters Applications of Differentiation
21. (a) V = x{s - Ixf, Q < X < -
dV
dx '
ItU - 2x)(-l) + {s- 2xY
= (s — 2x)(s — 6x) = Q when x = —,- {s/2 is not in the domain).
2 6
d'^V
^ = 24x - 85
dir
—-^ < 0 when x = —.
dx- 6
2s^ 5
V = — — is maximum when x = — .
27 6
(b) If the length is doubled, V = 27 (25)' = 8(575'). Volume is increased by a factor of 8.
23. 16 = 2y + ;c +
'(f)
12 = Ay + 2x + TTX
Zl -' Ix ~ TTX
A=^+2U
t^l x\2 (32 — 2x — TTX
\x +
( )
r
r
^r*
;;:::;::;:;::;::;:;:;;f:dii
2 4 8
^ = 8-x-f;c + ^;c = 8-
ctc 2 4
0 when x
8
xll.f
32
1 + (77/4) 4 + 77'
_=_^l+_j<Owhenx = ^^
_ 32 - 2[32/(4 + tt)] - 77(32/(4 + tt)]
16
4 + TT
16 32
The area is maximum when y = -:; — ; — feet and x = - — ; — feet.
4+77
4+77
25. (a)
y - 2 _^ 0 - 2
0 - 1 ~ .r - 1
y = 2 +
x- 1
L= Jx^ + y2
= /^
+ 2 +
;c - 1
;c^ + 4 +
X - 1 (x - 1)
;, X > 1
Z, is minimum when x = 2.587 and L == 4.162.
—CONTINUED-
Section 3. 7 Optimization Problems 149
25. —CONTINUED—
(c) Area = A{x) = -xy = -x\l + ^— -^
X + ■
X - 1
U - 1)2 = 1
X- 1 = ±1
;c = 0, 2 (select x = 2)
Then y = 4 and A = 4.
Vertices: (0, 0), (2, 0), (0, 4)
27. A = 2x>' = 2xV25 - x^ (see figure)
J 25 - 2r- \ „ , 572 ...
= 21 , _ 1 = 0 whenx = y = ^- = 3.54.
By the First Derivative Test, the inscribed rectangle of maximum area has vertices
,572 \/ 572 5^
~ 2 ' /'V~ 2 2
5 /5
Width: ^^; Length: 572
29. xy = 30 =^ y = —
A = {x + 2)[— + l] (see figure)
dA
dx
= (x + 2)1-^1 + I— + 2) = ^^^^^-T-^ = 0whenx = 730.
-W\ , /30
v2
. = ^=V30
730
By the First Derivative Test, the dimensions (x + 2) by {y + 2) are (2 +
7.411). These dimensions yield a minimum area
30) by (2 + 730) (approximately 7.477 by
31. V = Trr^h = 22 cubic inches or h =
22
(a)
Radius, r
Height
Surface Area
0.2
22
2-7HO.2)
.^•^ " .Ha2) J
= 220.3
77(0.2)2
0.4
22
tHO.4)-
2t7(0.4)
01
f\ 1 1
« 111.0
_°-' ' M0.4)=J
0.6
22
277<0.6)
->0
= 75.6
7t{0.6)-
L°-^ ' 77<0.6)2.
0.8
22
7T<0.8)2
27t(0.8)
11
= 59.0
L°-^ ' 7710.8) =
—CONTINUED—
150 Chapters Applications of Differentiation
31. —CONTINUED—
(b)
Radius, r
Height
Surface Area
0.2
22
277(0.2)[0.2 + J^^ J
== 220.3
0.4
22
277(0.4)
n^ 1 22 1
- 111.0
7T<0.4)2
°-^ ' 77(0.4)2]
0.6
22
277(0.6)[0.6 + ^2^2^^,
-75.6
77{0.6)2
0.8
22
77(0.8)2
277(0.8)[0.8 . ^^^^,
= 59.0
1.0
22
277(1.0)
'lO 1 22 "
= 50.3
77(1.0)2
.'•° ' 77(1.0)2_
1.2
22
277(1.2)
['■2 ^77(0)2] -45.7
41.2)2
1.4
22
277(1.4)
['■4 ^77(14)2] -43.7
77(1.4)2
1.6
22
277(1.6)
[^■^^77(iy-43.6
77(1.6)2
1.8
22
277(1.8)
b'uf.^K
= 44.8
77(1.8)2
2.0
22
77(2.0)2
27K2.0)
b-' ^ 4?0)2_
= 47.1
The minimum seems to be about 43.6 for r = 1.6.
33. Let X be the sides of the square ends and y the length of the package.
P = 4;c + y = 108 ^. .V = 108 - 4x
V = ;c2y = ;c2(108 - 4a:) = 108^2 - 4x^
dV
dx
= 216.x - 12a:2
12a:(18 - x) = Owhenx = 18.
22] ^ , 44
r + — r = 277r2 + —
The minimum seems to be 43.46 for r ~ 1.52.
(K AA
(e) — = 47rr - ^ = 0 when r = yil/77 «= 1.52 in.
dr r^
22
h= — r = 3.04 in.
7rr2
Note: Notice that
22
22
77r2 77(11/77)2/3 Vir'
= ^|i^)-^'.
d^V
dx"-
= 216 - 2Ax = -216 < Owhenx = 18.
The volume is maximum when x = 18 inches and y = 108 — 4(18) = 36 inches.
Section 3.7 Optimization Problems 151
35. V = -TTX^h = -Trx^{r + Vr^ - x^} (see figure)
3
cix~ 3''
2r^ + 2rVr2 - x^ - 3;c2 = 0
2rV/-2 - ;c2 = 3x^ - 2r^
0 = 9x^ - 8x-r2 = x\9x^ - Sr^)
2V2r
■ ^ = 0,^-
By the First Derivative Test, the volume is a maximum when
-. 4/-
X = — - — and h = r
(O.r)
■..-V7^)
+ Vr2 - x~
Thus, the maximum volume is
3277r3
-Hm
81
cubic units.
37. No, there is no minimum area. If the sides are x and y, then Ix -¥ 2y = 20 => y = 10 — .x.
The area is A{x) = jc(10 — x) = \Qx — x^. This can be made arbitrarily small by selecting x ~ 0.
39. V= 12 =-T7-7-3 + nr-h
12 - {4/3)T7r3 12 4
h = ^ = — J - -r
7rr- Trr'^ 3
12 4
S = Atrr- + lirrh = Airr^ + IttA — n - z;r
\Trr~ 3
24 8
-TTr~ +
24
— - = -irr T = 0 when r = IJ^Itt = 1.42 cm.
dr 3 r-^
— -r = -TT H — r > 0 when r = </9/Trcm.
dr- i r"
The surface area is minimum when r = 1/9/tt cm
and /! = 0. The resulting solid is a sphere of radius
r « 1.42 cm.
41. Let X be the length of a side of the square and y the length
of a side of the triangle.
4x + 3>' = 10
A 2^1 (^
A = x'- + -y\—y
(10 - 3v)' , V3 ,
• ■■ .•■; = 16 +— ^"
■•■ ■ | = i(io_3v)(-3)+^y^O
-30 + 9y + 4y3y = 0
30
■^ " 9 + 4v^
d-A 9 + 4^3
dy~
> 0
A is minimum when
30
y = 7= and .r = ~.
9 + 473 9 + 4V3
152 Chapters Applications of Differentiation
43. Let 5 be the strength and k the constant of proportionality.
Given /z^ + w- = 24^ h^ = 24^ - w^,
S = kwh^
S = kw(576 - w^) = k(576w - w^)
dS
dw
d'-S
= k(576 - 3w^) = 0 when w = 8^3, /i = 8V6.
= -6kw < Owhenw = SVS.
dw^
These values yield a maximum.
45. R = — sin 2d
8
dR 2v^ tt TiTT
—- = — ^cos 29 = 0 when 0 = — , —r-
dd g 4 4
^ = -^sin 29 < 0 when 6 = ^.
dd^ g 4
By the Second Derivative Test, R is maximum
when 8 = n/A.
47. sin a = - => i
s
, 0 < a < -
sm a 2
h , ^ 2 tan a ^
tano = -=>« = 2tana=>i = — : = 2 sec a
2 sm a
k sin a ksma k .
I = ;:; — = ; — = - sm a cos- a
s- 4 sec - a 4
-r- = T[sin a(—2 sin a cos a) + cos- a(cos a)]
da 4
= -rcos afcos^a - 2 sin- a]
4
= - cos a[l - 3 sin^ a]
„ , IT 377 , . 1
= 0 when a = —, -r-, or when sm a = ±—7= .
2' 2' V3
Since a is acute, we have
1
sm a
A = 2 tan a = 2 -7= = v^ feet.
73 \^l
Since (cm)/{doP) = {k/4) sin a(9 sin^a - 7) < 0 when sin a = 1/V3, this yields a maximum.
49.
s =
Jx'^ + 4,L
= 7i
+ (3 - .v)2
ne = ?- =
dT
y.x-2 + 4 Vx'~ -
2
- 6x -1- 10
4
;c- 3
dx
x^
2Vx2 + 4
9 - 6;c -^
4Vx-
x^
■ - 6a- + 10
ii:;t^
jc2 + 4 4(^2 - 6a -I- 10)
A^ - 6^3 -t- 9x2 + 8a - 12 = 0
You need to find the roots of this equation in the interval [0, 3]. By using a computer or graphics calculator, you can determine
that this equation has only one root in this interval (a = 1). Testing at this value and at the endpoints, you see that a = 1 yields
the minimum time. Thus, the man should row to a point 1 mile from the nearest point on the coast.
Section 3.7 Optimization Problems 153
51 7- = v^tM^ ^ V.r" - 6;c + 10
0^ ViV;c2 + 4 v.Va:^ - 6j: + 10
Since
X X — 3
= sin 0, and . ^ = — sin
V^^ + 4
we have
Vj:2 - 6x + 10
sin 9^ sin 62 sin 6, sin 62
Vi V2
Since
d^T 4
= 0 =
a[i2 v,U2 + 4)3/2 ^,^(^2 _ 5^ + 10)3/:
this condition yields a minimum time.
> 0
I
t ^
* £^
53. /W = 2 - 2sinx
3
2
I--
(a) Distance from origin to y-intercept is 2.
Distance from origin to j:-intercept is tt/2 ~ 1.57.
(b) d:
ix- + r
3
'x- + (2
: sin x)-
^
(0.7967. 0.9795)
Minimum distance = 0.9795 at.v = 0.7967.
(c) Let/U) = d\x) = ;c= + (2 - 2 sin.r)-.
f\x) = 2a: + 2(2 - 2sin;c)(-2cos.r)
Setting/'W = 0, you obtain .v == 0.7967, which
corresponds to d = 0.9795.
55. Fcose= k(W- Fsine)
^ kW
cos 6 + k sin 6
dF _ -kW{k cose- sine)
0
de (cos e + k sin 6)-
k cos e=sinfl=>A: = tane=>0 = arctan ^■
Since
1 k-
cos fl + ^- sin 0
the minimum force is
kW
Jk^+ 1 v//t= + 1
ftW
= Jk~ + 1,
cos e + /t sine Vit^ + 1'
1 «-J
154 Chapters Applications of Differentiation
57. (a)
Base 1
Base 2
Altitude
Area
8
8+16 cos 10°
8 sin 10°
-22.1
8
8+16 cos 20°
8 sin 20°
= 42.5
8
8 + 16 cos 30°
8 sin 30°
= 59.7
8
8 + 16 cos 40°
8 sin 40°
= 72.7
8
8 + 16 cos 50°
8 sin 50°
= 80.5
8
8+16 cos 60°
8 sin 60°
= 83.1
(b)
(c) A = {a + b)-
, 8 sin e
[8 + (8 + 16 cos e)]- ^
64(1 + cos e)sin e, 0° < 9 < 90°
(e) 100
Base 1
Base 2
Altitude
Area
8
8 + 16 cos 10°
8 sin 10°
= 22.1
8
8 + 16 cos 20°
8 sin 20°
= 42.5
8
8 + 16 cos 30°
8 sin 30°
= 59.7
8
8 + 16 cos 40°
8 sin 40°
= 72.7
8
8 + 16 cos 50°
8 sin 50°
= 80.5
8
8 + 16 cos 60°
8 sin 60°
= 83.1
8
8+16 cos 70°
8 sin 70°
= 80.7
8
8 + 16 cos 80°
8 sin 80°
= 74.0
8
8 + 16 cos 90°
8 sin 90°
= 64.0
The maximum cross-sectional area is approximately
83.1 square feet.
dA
(d) — = 64(1 + cos e)cos e + (-64 sin e)sin 6
du
= 64(cos e + cos^ d - sitP- 6)
= 64(2cos2e + cos e- 1)
= 64(2 cose- l)(cose+ 1)
= 0 when e = 60M80°, 300°.
The maximum occurs when 6 = 60°.
59. C = 1001
C'= 100'
200
x + 30
1 < X
400
30
(x + 30)2
Approximation: x ~ 40.45 units, or 4045 units
61. 5, = {Am - 1)2 + (5m - 6)^ + (10m - 3)^
- = 2(4m - 1)(4) + 2(5m - 6)(5) + 2(10m - 3)(10) = 282m - 128 = 0 whenm = -^.
dS__
dm
Line: y
S =
64
i4r
/^
\141
256
141
- 1
+
320
141
m
- 6
- 6
640
141
10
64
141
858
141
6.1 mi
^, o |4m - 11 |5m - 61 1 10m - 3
63. 5, = I , , ' + ' , _ ' +
Vm2+ 1 Vm2+ 1
m2+ 1
Using a graphing utility, you can see that the minimum occurs when x = 0.3.
Line: y = 0.3jc
|4(0.3) - 1| + |5(0.3) - 6| + 1 10(0.3) - 3|
^3 =
7(0.3)2 + 1
= 4.5 mi.
Section 3.8 Newton's Method
155
Section 3.8 Newton's Method
1. fix) =x^-3
fix) = 2x
X, = 1.7
n
x„
fiXn)
f'ixj
f'W)
X f^'^y
" nx„)
1
1.7000
-0.1100
3.4000
-0.0324
1.7324
2
1.7324
0.0012
3.4648
0.0003
1.7321
3. f(x) = sinj;
f'(x) = COS.X
;cj = 3
n
Xn
n^n)
/'UJ
■" /'UJ
1
3.0000
0.1411
-0.9900
-0.1425
3.1425
2
3.1425
-0.0009
-1.0000
0.0009
3.1416
5. f{x)=x^ +x- 1
fix) = 3x2+1
Approximation of the zero of/ is 0.682.
n
Xn
fi^n)
n\)
,fUJ
X ^^'"^
" f'(-\)
1
0.5000
-0.3750
1.7500
-0.2143
0.7143
2
0.7143
0.0788
2.5307
0.0311
0.6832
3
0.6832
0.0021
2.4003
0.0009
0.6823
/U) =
/'U) =
3Vx — 1 — X
3
2v^
Approximation of the zero of/ is 1.146.
Similarly, the other zero is approximately 7.854.
, /W = A^ + 3
fix) = 3.t2
Approximation of the zero of/ is - 1.442.
11. fix) = x' - 3.9^2 + 4.79x - 1.881
/'W = 3x' - 7.8x + 4.79
n
.r„
/UJ
n\)
/UJ
X ^^-'"^
" /'UJ
1
1.2000
0.1416
2.3541
0.0602
1.1398
2
1.1398
-0.0181
3.0118
-0.0060
1.1458
3
1.1458
-0.0003
2.9284
-0.0001
1.1459
n
•^„
fi^J
n^n)
/UJ
/'UJ
'" /'UJ
1
-1.5000
-0.3750
6.7500
-0.0556
-1.4444
2
-1.4444
-0.0134
6.2589
-0.0021
-1.4423
3
-1.4423
-0.0003
6.2407
-0.0001
-1.4422
n
■^n
/UJ
/'UJ
/UJ
/'UJ
, /UJ
" /'UJ
1
0.5000
-0.3360
1.6400
-0.2049
0.7049
2
0.7049
-0.0921
0.7824
-0.1177
0.8226
3
0.8226
-0.0231
0.4037
-0.0573
0.8799
4
0.8799
-0.0045
0.2495
-0.0181
0.8980
5
0.8980
-0.0004
0.2048
-0.0020
0.9000
6
0.9000
0.0000
0.2000
0.0000
0.9000
Approximation of the zero of/ is 0.900.
-CONTINUED—
156 Chapters Applications of Differentiation
11. —CONTINUED—
n
x„
no
/'UJ
nx„)
/'UJ
X f^'"^
" nx„)
1
1.1
0.0000
-0.1600
-0.0000
1.1000
Approximation of the zero of/ is 1.100.
n
Xn
nxj
/'UJ
/'UJ
" /'UJ
1
1.9
0.0000
0.8000
0.0000
1.9000
Approximation of the zero of/ is 1.900.
13. /W = .r + sin(.x + 1)
f'(x) = 1 + cos(;c + 1)
Approximation of the zero of/ is -0.489.
n
^.
nxj
/'UJ
nx,}
fixj
X ^^'"^
" fixj
1
-0.5000
-0.0206
1.8776
-0.0110
-0.4890
2
-0.4890
0.0000
1.8723
0.0000
-0.4890
15. h(x) = fix) - g(x) = 2;c + 1 - ^x + 4
1
h'(x) = 2
Ijx + 4
Point of intersection of the graphs of /and g occurs
when X = 0.569.
«
^„
h{x„)
hXx^)
X ^^'"^
1
0.6000
0.0552
1.7669
0.0313
0.5687
2
0.5687
-0.0001
1.7661
0.0000
0.5687
17. /zU) = /(j:) - g(x) = jc - tan X ,
h'{x) = 1 — sec-^jc
Point of intersection of the graphs of/ and g occurs
when X = 4.493.
n
Xr,
h(xj
k'(x„)
hix,}
h'ixj
X ^^'"^
" h'ixj
1
4.5000
-0.1373
-21.5048
0.0064
4.4936
2
4.4936
-0.0039
-20.2271
0.0002
4.4934
19. f(x) = x^ - a = 0
fix) = 2x
v2 _ „
2x,.^ - x,^ + a _ Xj^ + a _ x,- a
2x, 2Xi 2 2x:
i
1
2
3
4
5
X!
2.0000
2.7500
2.6477
2.6458
2.6458
77 = 2.646
23 , ^ 3.,^ + 6
i
1
2
3
4
Xl
1.5000
1.5694
1.5651
1.5651
76" « 1.565
Section 3.8 Newton's Method 157
25. fix) = 1 + cos X
f'(x) = -sinx
Approximation of the zero: 3.141
n
^n
/UJ
/'UJ
" fix J
1
3.0000
0.0100
-0.1411
-0.0709
3.0709
2
3.0709
0.0025
-0.0706
-0.0354
3.1063
3
3.1063
0.0006
-0.0353
-0.0176
3.1239
4
3.1239
0.0002
-0.0177
-0.0088
3.1327
5
3.1327
0.0000
-0.0089
-0.0044
3.1371
6
3.1371
0.0000
-0.0045
-0.0022
3.1393
7
3.1393
0.0000
-0.0023
-0.0011
3.1404
8
3.1404
0.0000
-0.0012
-0.0006
3.1410
27. y = 2jc3 - 6x2 + 6;c - 1 = f(^^)
y' = 6x^ - 12x + 6 =f'{x)
x, = l
fix) = 0; tlierefore, the method fails.
n
Xn
/UJ
/'UJ
1
1
1
0
29. y = -x^ + 6x2 - 10^ + 6 = /(x)
y'= -3x2 + lit - 10 =/'(x)
^1
= 2
Xj
= 1
X3
= 2
X4
= 1
and so on
Fails to
converge
31. Answers will vary. See page 222.
Newton's Method uses tangent lines to approximate c such that /(c) = 0.
First, estimate an initial x, close to c (see graph).
Then determine Xj by Xj = x.
/u.
fW
Calculate a third estimate by Xj = Xj
fix,)
Continue this process until \x„ - x„ + 1 1 is within the desired accuracy.
Let x„ + 1 be the final approximation of c.
W)
33. Let g(x) = fix) — X = cos x — x
g 'ix) = — sin X — 1 .
The fixed point is approximately 0.74.
n
x„
8ix„)
sXx„)
S^x„)
g\x„)
" S\x„)
1
1.0000
-0.4597
-1.8415
0.2496
0.7504
2
0.7504
-0.0190
-1.6819
0.0113
0.7391
3
0.7391
0.0000
-1.6736
0.0000
0.7391
158 Chapters Applications of Differentiation
35. fix) = jc3 - 3;t2 + 3, f'{x) = 3;c2 - 6x
(a)
(c) .^i =
1
Continuing, the zero is 2.532.
(e) If the initial guess jc, is not "close to" the desired zero
of the function, the x-intercept of the tangent line may
approximate another zero of the function.
(b) X, = 1
#4 « 1.333
f(Xl)
Continuing, the zero is 1.347.
(d)
The j:-intercepts correspond to the values resulting
from the first iteration of Newton's Method.
37. /U) = - - a = 0
X
fix) = -3
_^l"/^ 2" = x„+ x„'^[y - a]=x„+x„' x/a = 2x„- x^a = x„{2 - oxj
39. fix) = X cos X, (0, tt)
fix) = —X sin x + cos X = 0
Letting F{x) = fix), we can use Newton's Method as follows.
[F'{x) = — 2 sin X + X cos x]
n
Xn
F(x„)
nx„)
Fix„)
Fix,)
" F\x„)
1
0.9000
-0.0834
-2.1261
0.0392
0.8608
2
0.8608
-0.0010
-2.0778
0.0005
0.8603
Approximation to the critical number: 0.860
Section 3.8 Newton's Method
159
41. >'=/U) = 4-^2, (1,0)
d^- 7U -iy + (y- 0)2 = VU - D' + (4 - x-r- = ^x*-lx--2x + 17
li is minimized when D = x* — 7x- - Zx + 17 is a minimum.
gix) = D' = 4x^ - Hx - 2
g'ix) = 12r= - 14
n
•'^n
gix„)
sK)
1
2.0000
2.0000
34.0000
0.0588
1.9412
2
3
1.9412
1.9385
0.0830
-0.0012
31.2191
31.0934
0.0027
0.0000
1.9385
1.9385
x- 1.939
Point closest to (1. 0) is = (1.939, 0.240).
43.
, ,. . . ^ Distance rowed Distance walked
Minimize: T = — z ^ 1-
Rate rowed
Rate walked
_ vU2 + 4 V.T- - 6.x:
10
4
x-3
1.939.0.240)
3jx- + 4 4jx- - 6x + 10
4xjx- - 6x + 10 = -3U - 3)7.^2 + 4
16x2(^2 - 6x + 10) = 9U - 3Hx^ + 4)
7.r^ - 421^ + 43a" + 216a- - 324 = 0
Let /(a) = 7x* - 41v3 + 43^2 + 216x - 324 and /'(a) = 28.r3 - 126^2 + 86.1; + 216. Since /(I) = - 100 and/(2) = 56, the
solution is in the interval (1, 2).
n
Xn
fi^n)
n^)
/(■vj
, /K)
" /'(-^J
I
1.7000
19.5887
135.6240
0.1444
1.5556
2
1.5556
-1.0480
150.2780
-0.0070
1.5626
3
1.5626
0.0014
49.5591
0.0000
1.5626
Approximation: x == 1.563 miles
45. 2,500,000 = -76.x^ + 4830.x~ - 320,000
76x^ - 4S30x- + 2,820,000 = 0
Let/(x) = 16x^ - 4830.r2 + 2,820,000
f'(x) = ll%x~ - 966aT.
From the graph, choose .t, = 40.
n
^n
/(.vj
/'(•v„ )
/UJ
, /UJ
" /'UJ
1
40.0000
-44000.0000
-21600.0000
2.0370
37.9630
2
37.9630
17157.6209
-38131.4039
-0.4500
38.4130
3
38.4130
780.0914
-34642.2263
-0.0225
38.4355
4
38.4355
2.6308
-34465.3435
-0.0001
38.4356
The zero occurs when x == 38.4356 which corresponds to $384,356.
160 Chapters Applications of Differentiation
47. False. Let/W = (jr - \)/{x — 1). .r = 1 is a discontinuity. It is not a zero oif{x). This statement would be true if
fix) = pix)/qix) is given in reduced form.
49. True
51. fix) =\^ - 3.t2 + \x
fix)
3 2
1= — v^
6x + :
Let .»:[ = 12.
Approximation: a; = 11.803
Section 3.9 Differentials
n
-^.
/UJ
fK)
/'UJ
" /'UJ
1
12.0000
7.0000
36.7500
0.1905
11.8095
2
11.8095
0.2151
34.4912
0.0062
11.8033
3
11.8033
0.0015
34.4186
0.0000
11.8033
1. fix) =X'
fix) = 2x
Tangent line at (2, 4): >- - /(2) = /'(2)(x - 2)
y - 4 = 4ix- 2)
y = 4x — 4
X
1.9
1.99
2
2.01
2.1
fix) = X-
3.6100
3.9601
4
4.0401
4.4100
rU) = 4a: - 4
3.6000
3.9600
4
4.0400
4.4000
3. fix) = ;c= -. .
fix) = 5x^
Tangent line at (2, 32): y - /(2) = f'il)ix - 2)
y - 32 = 80U - 2)
>> = 80.x - 128
.r
1.9
1.99
2
2.01
2.1
/U) = ^
24.7610
31.2080
32
32.8080
40.8410
rU) = 80.x: - 128
24.0000
31.2000
32
32.8000
40.0000
5. fix) = s\nx
fix) = cos X
Tangent line at (2, sin 2):
y-/(2)=/'(2)(;c-2)
3; - sin 2 = (cos 2)ix - 2)
y = (cos 2)(x - 2) + sin 2
X
1.9
1.99
2
2.01
2.1
fix) = sin a:
0.9463
0.9134
0.9093
0.9051
0.8632
Tix) = (cos 2)(.x -
- 2) + sin 2
0.9509
0.9135
0.9093
0.9051
0.8677
3 2
1 j: , X
l.y=fix) = Wj'ix
Ay=fix + ^x) -fix,
= /(2.1)-/(2)
= 0.6305
2,Ax = dx = 0.1
flfy=/'(.x)a!x
= /'(2)(0.1)
= 6(0.1) = 0.6
Section 3.9 Differentials 161
9. y = fix) = x^ + \,f'{x) = 4x\ x= - 1, Ax = A = 0.01
Ay = f(x + Ax)- f(x) dy = f'(x) dx
= /(-0.99)-/(-l) =/'(-l)(0.01)
= [(-0.99)" + 1] - [(- ir + 1] « -0.0394 = (-4)(0.01) = -0.04
11. y = 3x^ - 4
dy = 6.t dx
15. y = xj\ - .v2
dy = \x
vr
+ Vl - -v- dx
Zx-
vT^^
iv
.r + 1
13.
>' =
2x-
1
dy--
-
3 .
(2;c-
-l)^'^
17.
y --
= 2x~
• cot-x
dy--
--(2 +
2 cot X csc-
.t)di:
=
= (2 + 2 cot ;c + 2
cot^.
x)dx
,o 1 /677nr- 1
19. y = -C0s(-^-
dy = — IT sin — ^7: | tic
\ 2
21. (a)/(1.9) =/(2 - 0.1) -/(2) +/'(2)(-0.1)
= 1 + (1)(-0.1) = 0.9
(b) /(2.04) =/(2 + 0.04) «/(2) +/'(2)(0.04)
== 1 + (1)(0.04) = 1.04
23. (a) /(1.9) =/(2 - 0.1) «/(2) +/'(2)(-0.1)
« 1 + (-jK-O.l) = 1.05
(b) /(2.04) =/(2 + 0.04) «/(2) +/'(2)(0.04)
= 1 + (-5)(0.04) = 0.98
25. (a) g(2.93) = ^(3 - 0.07) « ^(3) + ^'(3)(-0.07) 27. (a) ^(2.93) = ^(3 - 0.07) « ^(3) + ^'(3)(-0.07)
- 8 + (-5)(-0.07) = 8.035 . = 8 + 0{-0.07) = 8
(b) g(3.1) = ^(3 + 0.1) « g(3) + g'{3){0.\) . (b) g{2.\) = g{3 + 0.1) » ^(3) + g'(3)(0.1)
= 8 + (-5)(0.1) = 7.95 = 8 + 0(0.1) = 8
29. A=x-
x=\l
Ax = dx = ±^
dA = Ixdx
AA - (M = 2(12)(±^)
= ±f square inches
31. A = 7rr2 :
r= 14
Ar = dr = ±\
AA -^ dA = iTTr dr = 7r(28)(±5)
= ±777 square inches
162 Chapters Applications of Differentiation
33. (a) X = 15 centimeter
Ax = dx = ±0.05 centimeters
A=x^
dA = 2xdx = 2{15)(±0.05) .
= ± 1.5 square centimeters
Percentage error:
T = Bi = »■"«'■ ■ = ;'
W'^-^--^ 0.025
A x^ X
dx , 0.025
<
X ' 1
= 0.0125 = 1.25%
35. r = 6 inches
Ar = dr = ±0.02 inclies
4
dV = Anr'^dr = 477<6)2(±0.02) = ±2.887r cubic inches
(b) 5 = 4irr2
dS = 87rrrfr= 8i7<6)(±0.02) = ±0.967r square inches
dV ^ Airr^ dr ^ 3dr
(c) Relative error: y ~ (4/T,)Trr^ ~ r
= 7(0.02) = 0.01 = 1%
0
Relative error:
dS Sirrdr 2dr
Airr^
2(0.02)
= 0.000666 .
37. V = TTrVi = 407rr2, r = 5 cm, /z = 40 cm, dr = 0.2 cm
AV « rfV = SOTrrrfr = 80tt<5)(0.2) = SOwcm^
39. (a) T = Ittjm
ai —
/Ug
Relative
error:
dT
(ndD/i
gjL/g)
: T
27rv
'L/g
dL
IL
= — (relative error in L)
= -(0.005) = 0.0025
jnn 1
Percentage error: —(100) = 0.25% = -%
(b) (0.0025)(3600)(24) = 216 seconds
= 3.6 minutes
41. e =
-- 26°45'= 26.75°
de = ±15' = ±0.25°
(a) h = 9.5 CSC 6
dh = -9.5 CSC e cot Odd
^= -cotede .
h
dh
h
= (cot 26.75°)(0.25°)
Converting to radians, (cot 0.4669)(0.0044)
== 0.0087 = 0.87% (in radians).
(b)
= cotede < 0.02
0.02 tan e
de ^ 0.02
d ~ e(cot0)
de _^ 0.02 tan 26.75°
e ~ 26.75°
0
_ 0.02 tan 0.4669
0.4669
= 0.0216 = 2.16% (in radians)
Review Exercises for Chapter 3 163
43. r = ^(sin2e)
Vo = 2200 ft/sec
0 changes from 10° to 11°
(2200)2
dr =
»='«(il
16
Tso
{cos 28) d6
de={\\ - 10)
180
_ (2200)2 /'20irV it
16 '^°Tl80/Vl80
4961 feet
= 4961 feet
47. Let/(x) = ifx,x = 625, dx= -\.
fix + At) -/(x) + f'(x) dx= i/i +
A-'J?
dx
fix + A.r) = t/624 = ^625 +
4(4/625}
j(-l)
= 5-- = 4.998
Using a calculator, ^624 = 4.9980.
51. In general, when Ax -^ 0, dy approaches Ay.
45. Let/(;c) = ^,x= I00,dx= -0.6.
fix + Ax)^fix)+f'{x)dx
= Vx H -pdx
l-fx
fix + Ax) = 7995
= VTOO + — ^(-0.6) = 9.97
2V100
Using a calculator: 7994 = 9.96995
49. Let/(x) = Vx,x = 4.ir = 0.02,/'(x) = l/(2v^).
Thffli
/(4.02)-/(4)+/'(4)di
74^02 - V4 + ^^(0.02) = 2 + ^(0.02).
53. True
55. True
Review Exercises for Chapter 3
1. A number c in the domain of/ is a critical number if /'(c) = 0 or/'
is undefined at c.
/'Wis 3.. /V)=o
undefined. '
3. g(x) = 2x + 5 cos X. [0, 2-77]
^'(.r) = 2 - 5 sin X
= 0 when sinx = |.
Critical numbers: x == 0.41, x = 2.73
Left endpoint: (0, 5)
Critical number: (0.41,5.41)
Critical number: (2.73, 0.88) Minimum
Right endpoint: (277,17.57) Maximum
18
V
/
16.28. 17j7y
_^.73, 0.881/
4
164 Chapters Applications of Differentiation
5. Yes. /(- 3) = /(2) = 0. /is continuous on [- 3, 2],
differentiable on (— 3, 2).
f'(x) = (x + 3)0x - 1) = 0 for X = |.
c = 3 satisfies /'(c) — 0.
/(l)=/(7) = 0
(b) /is not differentiable atx = 4.
9. /(x) = x2/3, 1 < X < 8
/'W = |x-'/3
/(fc)-/(a) 4- 1 _3
b- a 8-1 7
/'(c) = |c-/3 = I
'14\3 2744 ^^^,
11. /(x) — X — COS X, — — < X < —
J V / '2 2
/'(x) = 1 + sinx
/(fc)-/(a)_ (77/2) -(-77/2)
fc-a (77/2) - (-77/2)
/'(c) = 1 + sin c = 1
c = 0
= 1
13. /(x) = Ar2 + Sx + C
/'(x) = 2Ax: + B
/(xj) - /(x,) /l(x/ - x,2) + S(x3 - X,)
= A(xi + Xj) + S
/'(c) = 2Ac + B = A(xi + Xj) + B
2Ac = A(xj + Xj)
Xj + X2
c = - — r — = Midpoint of [x,, XjJ
15. /(x) = (x - l)2(x - 3)
/'(x) = (x - 1)2(1) + (x - 3)(2)(x - 1)
= (x - l)(3x - 7)
Critical numbers: x = 1 and x = t
Interval:
-00 ~< X < 1
1 < X < 1
3 < X < 00
Sign oif'(x):
fix) > 0
fix) < 0
fix) > 0
Conclusion:
Increasing
Decreasing
Increasing
17. h{x) = v^(x - 3) = x3/2 - Sx'/^
Domain: (0, 00)
;!'(x)=|xl/2-|x-'/2
= 3 V2(^ _ . 3(x - 1)
2"^ ^ ^ 2v^
Interval:
0 < X < 1
1 < X < 00
Sign of h '(x):
h'{x) < 0
h'{x) > 0
Conclusion:
Decreasing
Increasing
Critical number: x = 1
Review Exercises for Chapter 3 165
19. h{t)
h'[t)
i,4
- 8f
8 = 0 when t = 2.
Relative minimum: (2, - 12)
Test Interval : -
oo < f < 2
2 < t < oo
Signof/j'W:
h'(t) < 0
h '(t) > 0
Conclusion:
Decreasing
Increasing
21. y = - cos(12r) - - sin(12/)
V = y'= -4sin(12f) - 3cos(120
77 1
(a) When t = '^,y = 'z inch and v = >> ' = 4 inches/second.
(b) y' = -4sin(12f) - 3cos(12r) = Owhen
sm(12r) 3 ,,^, 3
— TTTTT = -7 => tan(12r) = --.
cos(12r) 4 4
3 4
Therefore, sin(12r) = -— and cos(12i) = -. The maximum displacement is
I I)-K-|)=^'"^*^■
(c) Period: ^ = f
Frequency:
1 _ 6
77/6 77
23. f{x) = X + cosx, G < X < 2tt
f\x) = 1 - sin.r
77 377
f"{x) = - cos .r = 0 when x
r 2
^ . ^. „ ■ (tT 77\ /377 377
Pomts of mflection: I — , — I, I — , —
Test Interval:
0 < .V < ^
77 377
3- < -t < Y
377
— < .r < 277
Sign of /"(.v):
f"{x) < 0
f"(x) > 0
fix) < 0
Conclusion:
Concave downward
Concave upward
Concave downward
V2
25. g(x) = 2x2(1 - x^)
g'ix) = -4x{2x^ — 1) Critical numbers: x = 0, ±
g"(x) = 4 - 24x2
g"(0) = 4 > 0 Relative minimum at (0. 0)
g'i +—/= = — 8 < 0 Relative maximums at ±^7=, —
V V2; V V^ 2
27.
(5.^5))
29. The first derivative is positive and the second derivative is
negative. The graph is increasing and is concave down.
166 Chapter 3 Applications of Differentiation
31. (a) D = 0.0034r* - 0.2352/3 + 4.9423/2 - 20.8641/ + 94.4025
(b) 3«9
(c) Maximum at (21.9, 319.5) (=1992)
Minimum at (2.6, 69.6) (« 1972)
(d) Outlays increasing at greatest rate at the point of inflection (9.8, 173.7) (= 1979)
33. lim
2r2
= lim
X ->oo 3jc2 + 5 JT -»o=i 3 + 5/x^ 3
5 cos X
35. lim = 0, since |5 cos j-l < 5.
JT^OO X
37. /!(x) =
2x + 3
X- 4
Discontinuity: x = 4
, 2x + 3
lim — —
.X -»oo X — 4
nml±m = 2
.CO 1 - {4/x
Vertical asymptote: x = 4
Horizontal asymptote: y = 2
39. f(x)
- 2
Discontinuity: x = 0
lim I 2| = -2
J— »oo \X
Vertical asymptote: x = 0
Horizontal asymptote: 3; = — 2
41. fix) =x^ +
243
Relative minimum: (3, 108)
Relative maximum: (—3, — 108)
200
Vertical asymptote: x = 0
43. fix) =
1
1 + 3x^
Relative minimum: (-0.155,-1.077)
Relative maximum: (2.155, 0.077)
Horizontal asymptote: y = 0
45. fix) = 4x-x'^ = xi4 - x)
Domain: (-00, 00); Range: (-00,4)
fix) = 4 - 2x = 0when;<: = 2.
fix) = -2
Therefore, (2, 4) is a relative maximum.
Intercepts: (0, 0), (4, 0)
Review Exercises for Chapter 3 167
47. f(x) = xVl6 - x2. Domain: [-4, 4], Range: [-8,8]
Domain: [-4,4]; Range: [-8,8]
f'(x) = 1^^ ^ = 0 when x = ±2 V2 and undefined when x
f"i
Vie -X-
2x{x^ - 24)
±4.
(16 - x=p/2
/'(-2V2) > 0
Therefore, (-2^2, —8) is a relative minimum.
/"(272) < 0
Therefore, [2^, 8) is a relative maximum.
Point of inflection: (0, 0)
Intercepts: (-4, 0), (0, 0), (4, 0)
Symmetry with respect to origin
. hVz i)
2V1, -8
(-4.0) J 1(4.0)
I I t I f I * I — H^-t
i -6 I -2 i \ 2 4 6 8
^ (0. 0)
49. fix) = U - D^U - 3)2
Domain: (—00, 00); Range: ( — 00,00)
11
f'(x) = (x - iy{x - 3)(5x - 11) = 0 whenjt = 1, y, 3.
ii+ye
f"{x) = 4(x - l)(5x2 - 22.x + 23) = Owhen.r = 1,
/"(3) > 0
Therefore, (3, 0) is a relative minimum.
4t) <"
Therefore, ( — , 1 is a relative maximum.
Pomts of inflection: (1, 0), ( ' ^ ~ ^^, 0.60 ), ( " t 0.46
Intercepts: (0, -9), (1, 0), (3, 0)
51. fix) = .r'/5(.v + 3)2/3
Domain: (—00, 00); Range: (—00, 00
x+ 1
/W =
(X + 3)'/3;c2/3
-2
x^'Hx + 3)^/3
= 0 when x = —I and undefined when x = —3,0.
is undefined when .x = 0,-3.
By the First Derivative Test (-3, 0) is a relative maximum and (- 1. - v'^) is
a relative minimum. (0, 0) is a point of inflection.
Intercepts: (-3,0), (0,0)
168 Chapters Applications of Differentiation
53. fix)
x+ 1
X - 1
Domain: (-oo, 1), (1, oo); Range: (-oo, 1), (1, oo)
fix) = -r—^- < 0 if j: ^ 1.
f'Xx) =
ix - ly
Horizontal asymptote: y = I
Vertical asymptote: x = I
Intercepts: (- 1, 0), (0, - 1)
55. fix)
1 + x^
Domain: (-oo, oo); Range: (0, 4]
fix) =
-8x
(1 + x^-y-
= 0 when jc = 0.
/W= (i+,2)3 =Owhen. = ±— .
/"(O) < 0
Tlierefore, (0, 4) is a relative maximum.
Points of inflection: (±V3/3, 3)
Intercept: (0, 4)
Symmetric to the y-axis
Horizontal asymptote: >> = 0
57. fix) = x^ + x + -
■' X
Domain: (— oo, 0), (0, oo); Range: (— oo, — 6], [6, oo)
.„ ^ ^ n . 4 3xf* + x^ - 4 ^ ,
/ ix) = ix- +1 2 = ^ = 0 when x = ±\.
8 6;c^ +
fix) = 6x + ^ 3
7t 0
/"(- 1) < 0
Therefore, (- 1, -6) is a relative maximum.
/"(I) > 0
Therefore, (I, 6) is a relative minimum.
Vertical asymptote: x = 0
Symmetric with respect to origin
(-I.-6)
r\\
(1.6)
Review Exercises for Chapter 3
169
59. fix) =\x^-9\
Domain: (-00, oo); Range: [0, oo)
2x(jip- - 9)
f'(x) = — p^ — -j— = 0 when jc = 0 and is undefined when x =
2(x^ - 9)
fix) = -r^ — TTT is undefined at x = ±3.
F - 9|
/"(O) < 0
Therefore, (0, 9) is a relative maximum.
Relative minima: (+3, 0)
Points of inflection: (±3,0)
Intercepts: (±3, 0), (0, 9)
Symmetric to the )'-axis
61. fix) = X + cos JC
Domain: [0, lir]; Range: [1, \ + 2tt]
fix) = 1 — sin.T > 0, /is increasing.
fix) = - cos j: = 0 when x = —, -r-.
J \ J 2' 2
Points of inflection: — > TT 1 1 ^;~' ~;~
\2 2/ \ 2 2
Intercept: (0, 1)
±3.
(2ir, 2ir + l). ^
2ir-
Z''
. (f.f)/
"
v
</
(0, t)'
^•f)
1 ' i ' 21 "
63. x- + 4r - 2v - 16v + 13 = 0
(a) (.x:= - 2jc + 1) + 4(v= - 4>' + 4) = - 13 + 1 + 16
ix - 1)2 + 4Cy - 2)2 = 4
ix - 1)2 , (y-2)2
1
The graph is an ellipse:
Maximum: (1,3)
Minimum: (1, 1)
(b) X' + 4f -Ix- 16y + 13
0
2;c + 8v
dx
i4 = o
dx
^i%y - 16) = 2 - 2x
2 - 2jc
1 -X
dl_
dx 8y - 16 4v - 8
The critical numbers are .r = 1 and y = 2. These correspond to the points (1, 1), (1. 3), (2, - 1), and (2, 3).
Hence, the maximum is (1, 3) and the minimum is (1, 1).
170 Chapters Applications of Differentiation
65. Let r = 0 at noon.
L = rf- = (100 - 12/)2 + (- lOt)- = 10,000 - 2400t + 244?^
^ = -2400 + 488f = 0 when r = ^ - 4.92 hr.
dt 61
Ship A at (40.98, 0); Ship B at (0, -49.18)
d^ = 10,000 - 2400f + 244t^
=» 4098.36 when t « 4.92 « 4:55 p.m..
d » 64 km
(0,0)
(100-12(.0)
;
^'-4 (100,0)
(0, -100
67. We have points (0, v), {x, 0), and (1, 8). Thus,
y - 8 0-8 Sx
m = = ■ 7 or y
0-1 X - 1 ' X- V
8.t V
Let/(j:) = L^ = x- +
X- 1
f\x) = 2x+ 128'
X - 1
U-1)
U - 1)2 J
= 0
64x .
Jf ~ 7 TTT = 0
x\(x — 1)' — 64] = 0 when jc = 0, 5 (minimum).
Vertices of triangle: (0, 0), (5, 0), (0, 10)
2 4 6 8 10
69. A = (Average of bases)(Height)
v
- + ^^35^ + 2sx -x\ ^ ,
j (.see iigurej
dA_}_
dx 4
; , '''^' ""' , + 73.2 + 2sx - x^
L V3s2 + 2sx - x^
l{ls-x)(s + x) ^ ,
= — , , = 0 when x = 2s.
4j3s^ + 2sx - X-
A is a m
aximum when x = 2s.
|V3i^ + 2j:
71. You can form a right triangle with vertices (0, 0), {x, 0) and (0, y).
Assume that the hypotenuse of length L passes through (4, 6).
y - 6 6-0 6x
m = 7 = -: or J
0-4 A- x
Let/(x) = l2 = Jc2 + >>2 = ;c2 +
-x - 4
6.t
f'(x) = 2x + 12
,x- 4
X
ix - 4)2
= 0
\x - 4/
jr[(j: - 4)3 - 144] = 0 when .t = 0or;c = 4+ ^144.
L == 14.05 feet
Review Exercises for Chapter 3 171
73.
i-,
CSC e = — or L, =6 CSC 6 (see figure)
6
csci ^-0j = -^otU = 9 csc(y - e
L = L, + L, = 6 CSC e + 9 CSC
e = 6 CSC e + 9 sec e
cf0
-6csc ecot e + 9 sec dlan 6 = 0
tan' 0 = - => tan
j/2
1/2
seed = Vl + tan= 6 = , / 1 +
/
2\2/3 V32/3 + 22/3
CSC 6 =
: e V3-/3 + 2-'
/3
Z. = 6
tan e 2'/3
(32/3 + 22^^ „ (32/3 + 22/3)'/^
L y
X 6
UyC
Xe '
\
/{e _
(!-«)
9
21/3
+ 9-
31/3
3'/3
= 3(32/3 + 22/3)3/2 ft == 21.07 ft (Compare to Exercise 72 using a = 9 and fc = 6.;
75. Total cost = (Cost f)er hour) (Number of hours)
^ /v2 ^VllO\ llv , 550
^=ii_ 550 ^ llv- - 33,000
dv ~ 60 V' 60v-
= 0 when v = ^3000 = 10^30 = 54.8 mph.
dv- ~ v3
> 0 when v = lOv^ so this value yields a minimum.
77. fix) = .v3 - 3x - 1
From the graph you can see that/(.r) has three real zeros.
f\x) = 3.x2 - 3
n
K
/(^„)
/'(^J
1
-1.5000
0.1250
3.7500
0.0333
-1.5333
2
-1.5333
-0.0049
4.0530
-0.0012
-1.5321
n
^n
/(■^J
/'(-^J
/(-T„)
X ^^'"^
" f'ix„)
1
-0.5000
0.3750
-2.2500
-0.1667
-0.3333
2
-0.3333
-0.0371
-2.6667
0.0139
-0.3472
3
-0.3472
-0.0003
-2.6384
0.0001
-0.3473
n
Xn
fix„)
/'(■vj
1
-1.9000
0.1590
7.8300
0,0203
1.8797
2
1.8797
0.0024
7.5998
0.0003
1.8794
The three real zeros of/(j:) are .r = - 1.532, x = -0.347, and x = 1.879.
172 Chapters Applications of Differentiation
79. Find the zeros of/(;c) = X* — x — 3.
fix) = 4;t^ - 1
From the graph you can see that/(j:) has two real zeros.
/changes sign in [—2, — l].
n
Xn
/UJ
f'iXn)
1
- 1.2000
0.2736
-7.9120
-0.0346
-1.1654
2
-1.1654
0.0100
-7.3312
-0.0014
-1.1640
On the interval [-2, -l]:x~ -1.164.
/changes sign in [1, 2].
n
x„
/UJ
/'UJ
1
1.5000
0.5625
12.5000
0.0450
1.4550
2
1.4550
0.0268
11.3211
0.0024
1.4526
3
1.4526
-0.0003
11.2602
0.0000
1.4526
On the interval [1, 2]; x = 1.453.
81. y = x{\ - cos j:) = x - j: cos j:
die
1 + j: sin^: — cos;c
dy = (1 + ;c sin j: — cos x) dx
83. 5 = 47rrl rfr = Ar = ±0.025
dS = S'nrdr= 877(9)(±0.025)
= ± 1. 8 TT square cm
f(100) = ^(100) = ^(100)
47rr
2(±0.025)
(100) = ±0.56%
4
dV = 4Trr2 dr = 47r(9)2(+0.025)
= ±8.1 77 cubic cm
3dr
f'><»> = iJ^«
3(±0.025)
(100)
(100)
rO.83%
Problem Solving for Chapter 3
1. Assume y^ < d < y^. Let g{x) = f{x) — (/(jc — a), g is continuous on [a, b] and therefore
has a minimum (c, g(c)) on [a, b]. The point c cannot be an endpoint of [a, b] because
g'{a)=f'(a} -d = y^-d<Q
g'ib) =f'{b) -d = y^~d>Q
Hence, a < c < b and g'{c) = 0 => f'{c) = d.
Problem Solving for Chapter 3 173
3. (a) For a = —3, —2, —\,0, p has a relative maximum at (0, 0).
For a = 1, 2, 3, p has a relative maximum at (0, 0) and 2 relative minima.
(b) p'{x) = Aax' - \lx = Ax{ax^ - 3) = 0
0.±
p"{x) = \2ax- - 12 = 12(a.v2 - 1)
For x = 0, p"(Q) = -12<0=>phasa relative maximum at (0, 0).
(c) If a > 0..r = ±
P ±
- are the remaining critical numbers.
= 12( — I — 12 = 24 > 0 => p has relative minima for a > 0.
a I \a
(d) (0, 0) lies on y = -Ix-.
^3
Let.t = ±
a
. Then
^(^)="(!r-<i . .
9 / /3\
Thus,>'= — = -3 ±^/- = -3j:- is satisfied by all the relative extrema of p.
a \ \ a I
5. p{x) = x"^ + ax^ + \
(a) p '(x) = 4:^ + lax = 2x(2.c- + a)
p"(x) = I6;c= + 2a
For a > 0, there is one relative minimum at (0, 0).
(b) For a < 0, there is a relative maximum at (0, 1).
(c) For a < 0, there are two relative minima at .v = ±
(d) There are either 1 or 3 critical points. The above analysis
shows that there cannot be exactly two relative extrema.
7. f(x} = - + .r2
X
fix) = — T + 2j: = 0 => -^ = 2v
X- x-
2 \ X
/"W = | + 2
If c = 0,f{x) = x-'^ has a relative minimum, but no relative maximum.
If c > 0, X = 3/ - is a relative minimum, because/"! -V - J > 0.
c .
If c < 0, .V = ..^V r is a relative minimum too.
Answer: all c.
174 Chapters Applications of Differentiation
9. s./^'^-f^f-f'lf'-^^ = k.
(b - ay
Define F{x) = f{x) - f(a) - f'{a)ix - a) - kix - af.
F{a) = 0, F{b) = fib) - f(a) - f'{a){b -a)- kib - a)^ = 0
F is continuous on [a, b] and differentiable on (a, b).
There exists c,, a < c, < fc, satisfying F'(c,) = 0.
f W = fix) — f'ia) — 2kix - a) satisfies the hypothesis of RoUe's Theorem on [a, c,]:
F'ia) = 0, F'(ci) = 0.
There exists Cj, a < C2 < c, satisfying F"ic2) = 0.
Finally, F"ix) = fix) - Ik and F'Xcj) = 0 implies that
Ti,„o ; /(fe) - fia) - fia)ib - a) f'jc^ _^ffu\ ft \ j_ t'( \(u ^ _■_ ^ ft \(u \2
Thus, k = _ .^ = —7^ =^fib) = fia) +fia)ib - a) + -f (c,)!^ - a)^.
, , tan 4>i\ - 0.1 tan (/)) 10 tan </> - tan- 0
■ ^'^' ~ 0.1 + tan(/) ~ 1 + 10tan</)
^,, J, (1 + lOtan (^)(10sec2<^ - 2tan<^sec2<^) - (lOtan (^ - tan2(^)lOsec-0 „
^^"^^^ (1 + 10 tan 4.) = °
=> (1 + 10 tan (^)(10 sec- 0 - 2 tan (^ sec^ <^) = ( 1 0 tan <^ - tan- <^)10 sec^ 0
=* 10 sec- (/) - 2 tan (^ sec-^ <^ + 100 tan <^ sec- (^ - 20 tan^ </> sec- <^
= 100 tan (^ sec* </> - 10 tan^ <^ sec- <^
=> 10 - 2 tan <^= 10 tan- 0
=> 10 tan^ (^ + 2 tan (^ - 10 = 0
— 2 -I- /4 + 400
tan (^ = =^^^- = 0.90499, - 1.10499
Using the positive value, <f> = 0.7356, or 42.14°.
13. V = -2400irsine
v'= -2400iTCOse = 0
t* = — + 2/jTr, — -(- 2«ir, n an mteger
Problem Solving for Chapter 3 175
X y 4
15. The line has equation T + T = 1 or y = --x + 4.
Rectangle:
Area = A = xy = x[ --x + 4J = ~:;x- + 4x.
A '{x) = -|.r + 4 = 0=>|x = 4
Dimensions: r x 2 Calculus was helpful.
X V
Circle: The distance from the center (r, r) to the line T "*" J ~ ' ~ ^ must be r.
fn-
- 1
12
7r- 12
/l
1
5
12
|7r- 12
5r = |7r - 12| =^ r = 1 or r = 6.
Clearly, r =1.
X y
Semicircle: The center lies on the line T + T = ^ ^^ satisfies x = y = r.
Thusf + ^=l=>^r=l
r = —. No calculus necessary.
17. >> = (1 + j:')-'
^ (1 + x-y-
" 2(3a-^ - 1) . _ ^1 ^V3
/': 1 1 h-
3 3
The tangent line has greatest slope at ( — — , - I and least slope at . ,
V3 3
73 3
19. (a)
.V
0.1
0.2
0.3
0.4
0.5
1.0
sinj:
0.09983
0.19867
0.29552
0.38942
0.47943
0.84147
sinj: < .r
(b) Let f(x) = sin x. Then f'(x) - cos x and on [0, x] you have
by the Mean Value Theorem,
f,, . fix) -no) „
f(c) = ^_o ' 0 < c <.v
cos(c)
Hence,
X
sin.r
|cos(c)| < 1
|sinA:| < |jc|
sin.v < .V
CHAPTER 4
Integration
Section 4.1 Antiderivatives and Indefinite Integration 177
Section 4.2 Area 182
Section 4.3 Riemann Sums and Definite Integrals 188
Section 4.4 The Fundamental Theorem of Calculus 192
Section 4.5 Integration by Substitution 197
Section 4.6 Numerical Integration 204
Review Exercises 209
Problem Solving 214
CHAPTER 4
Integration
Section 4.1 Antiderivatives and Indefinite Integration
Solutions to Odd-Numbered Exercises
1. Ml ^A= Aj-x^-i ^r\^ _Q.-4 -9
M.^^^j^^^^"^^^ = -^'"
d n
3. ^1 -x3 - 4x + C] = X- - 4 = {x - 2)ix + 2)
5.f = 3r^
dt
7.$; = x3/2
y = fi + C
3' = \x?n + C
Check: -^^^
at
+ C] = 3/2
Check: £ Ir^/^ + C
Given
Rewrite
Inteerate
Simplify
9. 3/^aL^:
x"^ dx
_^4/3
4
11. -^dx
J xwx
x-^'^-dx
^-
~i*"
13./i<i.
- ip"^
^S)-
-i-
15. {x + 3)dx =
y + 3;t + C
17. (Zr - 3.r-)ciT: = x' -
Check:
dx
- + 3. + C
= ;^/2
= .v + 3
Check: —[;<:-- .r^ + C] = 2x - 3.v=
/<
19. I (x' + 2) (ir = -r'' + 2jr + C
Check: jijx* + 2.x + C| = .v^ + 2
/'
21. I (x3/2 + 2x + 1) a[r = t^v^''^ + ,.; + ^ + ^
Check: M^'' + x- + x + c\ = x^'- + Iv + 1
23. \l/^dx= j.r
jc^''' 3
Check: 4(|^-'^' + C) = .x^/s = l/^-
25. 3 ^^ = -^^^ ^^ = 3T + C =
Check: ^(--i^ + C=^
dx\ Ix^ j .r"
-^ + c
177
178 Chapter 4 Integration
27. {^^^j^dx = I (;c3/2 + x'/'- + x-'/2) dx = '^x?'^ + |jc3/2 + 2x'/2 + C = -^;c'/2(3x2 + 5x + 15) + C
15
Check: -f fl-^'' + 1^^' + 2x"' + c) = i^'^ + x'/^ + ;c-'/2 = ""' "^ ^^"^ '
dx\5 3 / y^
'■/
/<
29. U + l)(3x - 2) dlx = (3x2 + ;c - 2) ^^
= ^3 + -x2 - 2x + C
2
Check: 4-\^ + ~:^ - 2x + C] = 3x^ + x - 2
dx\
= ix+ l)(3x - 2)
31. J/vS^y = J;
//2 dy = -yin + C
Check:
dj2
dy
±yin + cj = //2 = y2^
33. Uf =
\dx = x + C
\
35. I (2 sin X + 3 cos x) dx = — 2 cos x + 3 sin x + C
Check: — (x + C) = 1
OK
Check: -7-(— 2 cos x + 3 sin x + C) = 2 sin x + 3 cos x
ox
/
37. (1 - CSC t cot i)dt = t -\- esc r + C
\
39. (sec^ e - sin 0) dS = tan e + cos e + C
Check: —it + esc r + C) = 1 - esc r cot r
Check: —(tan e + cos 6 + C) = sec^ 0 - sin 0
dv
1. (tan-y + \)dy = \.
41. (tan- y + \)dy = sec' >> dy = tan y + C
Check: —(tan y + C) = sec- y = tan- y + \
dy
43. /(x) = cos X
C = -2
45. /'(x) = 2
/(x) = 2x + C
f(x) = It + 2
/W = Tx
Answers will vary.
49. ^ = 2x - 1, (1, 1)
dx
-\
(Ix - \)dx = x^-x + C
1 = (1)2- (1) -t- C => C= 1
>> = x^ — X -1- 1
Answers will vary.
Section 4.] Antiderivatives and Indefinite Integration 179
51. V- = cos X, (0, 4)
dx
53. (a) Answers will vary.
^h
y = I cos X dx = s'mx + C
4 = smO + C=> C = 4
y = sin.ic + 4
(b)J = 4 -1,(4,2)
dx 2
y=--x+C
42
2 = — -4 + C
4
2 = C
55. fix) = 4x,f(0) = 6
f(x) = 4r ^ = 2i2 + C
y{0) = 6 = 2(0)2 + c => C = 6
/(;c) = 2;c2 + 6
57. h'{t) = 8f3 + 5, /!(]) = -4
;i(f) = (8r3 + 5)dt = 2r* + 5f + C
/!(1) =-4 = 2 + 5 + C^C = -11
/!(?) = 2f^ + 5f - 11
59. f"{x) = 2
/'(2) = 5
/(2) = 10
/W = I 2 dx = 2j: + Ci
/'(2) = 4 + C, =5=>C, = 1
/'(;t) = 2x + 1
fix)= \{2x+ l)dx = x- + x+ Cj
/(2) = 6 + Q = 10 =^ Q = 4
/(x) = jr + .r + 4
= /
63. (a) h(t) = {l.5t + 5)dt = 0.75r + 5t + C
h{0) = 0 + 0 + C=12=i.C=12
/!(;) = 0.75r + 5f + 12
(b) ;!(6) = 0.75(6)2 + 5(6) + 12 = 69 cm
61. f"(x) = x-3/2
/'(4) = 2
/(O) = 0
fix
\x~^/-dx = -2r-i/2 + C, = -- %r + C,
/'(4) = -- + C, = 2 ^ C, = 3
/'(x) = -^ + 3
/(;c)
(-2r-'/2 + 3) iv = -4.v"- + 3.r + C,
/(O) = 0 + 0 + C, = 0=* C, = 0
fix) = -4x^'- + 3.T = -4^^ + 3.V
180 Chapter 4 Integration
65. /(O) = - 4. Graph of/' is given.
(a)/'(4)«-1.0
(b) No. The slopes of the tangent lines are greater than 2
on [0, 2]. Therefore, /must increase more than 4 units
on [0, 4].
(c) No,/(5) < /(4) because /is decreasing on [4, 5].
(d) /is an maximum at x = 3.5 because /'(3. 5) = 0 and
the first derivative test.
(e) /is concave upward when/' is increasing on (-oo, 1)
and (5, oo)./is concave downward on (1, 5). Points
of inflection at j: = 1,5.
67. ait) = -32ft/sec2
v(f) = \ -32 dt = -32t+ Ci
v(0) = 60 = C,
s{t) = (-32f + 60)dt = - \6t- + 60f + C^
5(0) = 6 = C.
5(r) = -16r^ + 60f + 6 Position function
The ball reaches its maximim height when
v(t) = -32r + 60 = 0
32? = 60 ■ ' ■
15
t = — seconds
o ' - .
15
161 -^ I +601
(f
+ 6 = 62.26 feet
asy
71. a(f) = -9.8
v(r) = -9.8 ^f= -9.8<+ C, .
v(0) = Vq = C, ^ v(f) = -9.8r + vo
/(/) = (-9.8r + Vo) dt = -4.9t^ + v^t + C^
m = 5o = Q ^ fit) = -4.9f2 + v^t + So
(f) /"is a minimum at x = 3.
(g)
69. From Exercise 68, we have:
sit) = -16r2 + V(,f
s '(f) = - 32f + Vji = 0 when r = — = time to reach
maximum height.
it2)—''[f2r'it2)^'''
'°' + ^ = 550
64 32
35,200
187.617 ft/sec
73. From Exercise 71, /(f) = -4.9t^ + lOf + 2.
V (f) = -9.8f + 10 = 0 (Maximum height when v = 0.)
9.8f = 10
10
t =
9.8
/I
©=■
7.1m
75. a = -1.6
- v(f) = — 1.6 (if = — 1 .6f + Vq = — 1.6f, since the stone was dropped, Vq = 0.
sit) = (-1.6f)rff = -0.8f2 + 5o
j(20) = 0 => -0.8(20)2 + So = 0
5o = 320
Thus, the height of the cliff is 320 meters.
v(f)=-1.6f
v(20) = -32 m/sec
Section 4.1 Antiderivatives and Indefinite Integration
181
77. x{t) = r3 - 6f2 + 9f - 2 0 < r < 5
(a) vW = x'(t) = 3f2- 12r + 9
= 3(r= - 4f + 3) = 3(f - l)(f - 3)
a(t) = v'W = 6f- 12 = 6(r- 2)
(b) v(f) > 0 when 0 < r < 1 or 3 < r < 5.
(c) a(t) = 6(r - 2) = 0 when t = 2.
v(2) = 3(l)(-l) = -3
79. v{t) = -^ = r'/2
f > 0
xit)
= /.
(f) dt = 2f'/= + C
x{\) = 4 = 2(1) + C => C= 2
jc(r) = 2t'^' + 2 position function
a(r) = V '(r) = - r' ^^^ = TI72 acceleration
81. (a) v(0) = 25 km/hr = 25 • j^ = — m/sec
v(13) = 80km/hr = 80-|^ = ^m/sec
a(f) = (3 (constant acceleration)
v{t) =at+C
v(0) =
v(13) =
250
36
800
36
550
36
250
v(r) = «r + —
13a +
13a
250
36
550 275 , ,^^ , ,
'' = 468=l3i'='-''^'"/^^'^"
(b) ^W = aY + ^f Wo) = o)
275 (13)- 250
„ (lmi/hr)(5280 ft/mi) 22
**^- (3600sec/hr) " 15 "'^^'^
(a)
83. Truck: v{t) = 30
. s{t) = 30f (Let 5(0) = 0.)
Automobile: a{t) = 6
v(/) = 6f (Let v(0) = 0.)
5(f) = 3/2 (Let s(<S) = 0.)
At the point where the automobile overtakes the truck:
30f = 3r=
0 = 3r - 30r
0 = 3r(r - 10) when t = 10 sec.
(a) 410) = 3(10)- = 300 ft
(b) v(10) = 6(10) = 60 ft/sec = 41 mph
;
0
5
10
15
20
25
30
V, (ft/sec)
0
3.67
10.27
23.47
42.53
66
95.33
V,(ft/sec)
0
30.8
55.73
74.8
88
93.87
95.33
,c)5,(,)./v,(,).,.5Jp^-!if^^ + 0,3679,
= /
5,(f) = V,(f) dt
0.1208r3 6.7991/2
(b) V(f) = 0.1068/2 - 0.0416/ + 0.3679
VjCr) = -0.1208/2 4. 6.7991/ - 0.0707
0.0707/
3 2
[In both cases, the constant of integration is 0 because 5,(0) = Sjfi) = 0]
5,(30) = 953.5 feet
52(30) «= 1970.3 feet
The second car was gomg faster than the first until the end.
182 Chapter 4 Integration
87. a(t) = k
v{t) = kt
s(t) = ^P since v(0) = s(0) = 0.
At the time of lift-off, kt = 160 and {k/2)f = 0.7. Since {k/l)fi = 0.7,
t =
¥-V¥=«
\Ak = 1602
1602
1.4
^ 18,285.714 mi/hr2
= 7.45 ft/sec^.
89. True
91. True
93. False. For example, Ix-xdxi' Ixdx- I.
X dx because — + C ^t ( y + C, jfy + Cj
95. fix)
1, 0 < .r < 2
3x, 2 < .x: < 5
I .t + Ci, 0 < .r < 2
ll" "^ ''2'
/(I) = 3 =^ 1 + Ci = 3 =i> C, = 2
/ is continuous; Values must agree ?& x = 2:
4 = 6 + C2=>C2 = -2
fjc + 2, 0 < ;r < 2
fix) = 3a:2 ^ ^
— -2, 2<x<5
The left and right hand derivatives at .t = 2 do not agree. Hence / is not differentiable at x = 2.
Section 4.2 Area
5 5 5
1. ^(2/ + 1) = 2^/ +^1= 2(1 +2 + 3 + 4 + 5) + 5 = 35
i=\ i=\
3 Y '
-1 + 1 + 1+1 + 1 158
9 1
" ' ^ 2 ^ 5 ^ 10 + 17
8
'■2.
85
15.f2, = 2f,^2M=420
1=1 /=i L -i J
5. ^c = c + c + c + c = 4c
20 19
17. ;^(i-i)2=2r-
1=1 >=i
'19(2(
))(39)
-iiK-fn
L 6 J
2470
Section 4.2 Area 183
15
2
;=1
15
19. 2 '■(/- 1)^ = !:'■'- 21 '' + E
= 1 i=l
^ \5\\6Y _ .15(16)(31) 15(16)
4 6 2
= 14,400 - 2.480 + 120
= 12,040
21. sum seqUB 2 + 3, ;c, 1, 20, 1) = 2930 (n-S2)
2(r- + 3) = 20i2o±iM20LLi)^3,2o,
20
i=l
(20)(21)(41)
60 = 2930
23. 5 = [3 + 4 + f + 5](1) = f = 16.5
i = [1 + 3 + 4 + f](l) = f = 12.5
25. 5 = [3 + 3 + 5](1) = 11
.y = [2 + 2 + 3](1) = 7
- «^) = J^ - Vf (^ M) ^ 4^ = '^-^;-^^^ = «-™
'<^. - G)* VKa - VK^
3/l\ 1 + 72+^/3
4V4
« 0.518
M-M4—
i.l.i.i.i»o.«.
31. lim
8lV^(«+lFl 81.
~r : = ~r lim
/j''/ 4 J 4n-»oo
n"* + 2n3 + n-
>-7
33. lim
n— >oo
18VH" + 1)
18,.
— hm
n- + n
> = 9
35. ;^ii4^ = -L2(2< + i) = A
5(10) = ^ = 1.2
- 5(100) = 1.02
5(1000) = 1.002
5(10,000) = 1.0002
.n{n + 1)
2 1- n
n + 2
■-Sin)
37-1
^6k(k - 1) 6 ^
^^(k^-k)
n{n + 1)(2« + 1) n{n + 1)
2«- + 3n + 1 - 3« - 3
[2n- - 2] = 5(n)
5(10) = 1.98
5(100) = 1.9998
5(1000) = 1.999998
5(10,000) = 1.99999998
,„ ,• v^/16'\ ,. 16 ^. ,. 16/r7(« + D'
39. hm y -T = lim — V/ = lim — P-r — -\ = Urn
n-><x> l^^\ n'- 1 n->oo n- fr'^ n->Qo n-\ 2
m
8 lim 1 +- = 8
n^^ooV /I/
184 Chapter 4 Integration
41. lim > -t(i - 1)^ = lim ^ > i^ = lim -^\
n->co OL « J n-'ooL OV 1
43. lim y{\+ -\{-\ = 2 lim -[y 1 + i y il = 2 lim -
l/n(n+l)
= 2 lim [1+4:^
i-»oo L 2n^
211+2
45. (a)
(b) Ax
2-0 2
n n
Endpoints:
o<,g)<.g)<.^.<„-„g)<„g) = .
(c) Since >> = j: is increasing, /(m,) = /U,._ J on [;c,-_ ,, x,].
^(n) = J/U,_,)Ax
(2\ '
(d)/(W,)=/(j:,-)on[A:,-_„x,]
(<■ - 1)
*.=|/«-=,|/(!M,[.M!)
47. >> = -2x + 3 on [0, 1]. (Note: \x = ^-^^ = -
V n n
= 3
2 ^ . , 2(n + l)n „ 1
lE' = 3
2«2
= 2
Area = lim s{n) = 2
49. 3^ = ^2 + 2 on [0, 1]. (Note: Ax = -
-"Hm-t
+ 2
[il'
(e)
X
5
10
50
100
s{n)
1.6
1.8
1.96
1.98
S{n)
2,4
2.2
2.04
2.02
(f) lim 2
(<• - 1)
^^= iim^|;a-i)
«/ n^oo n^
.. 4r«(«+_i) ]
= hm — n
n->OG n\ 2 J
= lim r^(^l±il - i] = 2
/!->00 L " ''J
4 \«(n + 1)
= lim
= lim2i^±i) = 2
3 1
^ n(n + 1)(2« +1) ^ ,^ _ . , ,
+ 2 = -5^ 7^\ ^ + 2 = -2+- + — +2
6n^ 6\ n n-
H 1 — -^i
Area = lim S{n) = —
n-»oo 3
Section 4.2 Area 185
51. y = 16 - A-2 on [l, 3]. ( Note: Ax = -
«(.)^|/(. -!)© = ,!
9 "
1
r
\t>n
n
,
15
4r 4i'
16 - I 1 + —
4 n{n + l)(2;i + 1) 4 «(>; + 1)
«"
30 - -^(/! + l)(2n + 1) - -(/I + 1)
6/r /!
53. y = 64 - .x^ on [1, 4]. (Note: Ax
^(«) = |/ii+7
n /\n
64 - 1 +
3(
'' i= 1
63
27i-' 27r 9i
63n
27 n^n + lY 27 >;(>; + 1)(2k + 1) 9 n(n + 1)'
= 189-fi(.+ l)-fi,(n.l)(2..1)-?^i^
4n- 6/!- 2 /!
81 27 513
Area = lim s{n) = 189 - ^ - 27 - -^ = — - = 128.25
n ->ao 4 2 4
55. y = -t^ - x3 on [- 1. 1]. Note: A.^
l-(-l)
Again, T{n) is neither an upper nor a lower sum,
2i\/2
1 = 1
1=1
n
= 1
1+^
■1 +
7;\3'
, 4( 4/-
1 -- + —
n n~
, 6i 12/2 8,-3
■1 + - + —
n n- n-
= 2
i=l
20/ 26i: _ sii
/I H- n'
4^
20A . 32,^
16^
nl n
-Si-^I'^^I^^-^Z''
n
n ;
4, , 20 n[n +1) 32 n(" + lH2'! +1) 16 ^!-(/i + 1)-
-(") — - ■ — ~ + -J • 7 T ■ ;
n n- 2 If' 6 ;;* 4
4-10|l+iUi^2 + ^ + A
41 1 + - + -
;; II-
32 2
Area = lim T{ii) = 4- 10 + ^-4 = -
n— >cc 3 3
186 Chapter 4 Integration
2-0 2
57. /( v) = 3y, 0 < .V < 2 Note: ^y =
«.=i/w.,=|/(f)e)=|3(f)e
12 A. /12\ n(n + 1) 6(n + 1) ^ 6
= 711'= -I- -^— = = 6 +
Area = lim S{n) = lim ( 6 + - 1 = 6
59. /(y) =r, 0 <y < 3 (Note: Ay
3-0 3
«")vS/(!B = S(!)B=5i.-
nj \ni n
^ Tl n(n + \){2n + 1) ^ 9 l2n- + ^in + \\ ^27 9
" «3 ■ 6 " nA 2 / 2n 2n2
Area = lim S{n) = lim ( 9 + ^ + -^ ) = 9
n-»oo 'i-»oo V 2« 2«^/
61. gW
/ 3-12
= 4y2 - y3, 1 < V < 3. Notc: Ay = = -
V n n
sw = 2^ 1 +
2i\/2
= E
4 1 +
n /\n
2[
n
1 +
-)1-
2 " r, 4i API r, 6( 12r 8/3
-y41+ — + — r- -1+ — + —z- + —r
1 1^1 L n n^ ] L n n- n _
2^r3,M^4._8.1
10 n{n + 1) ^ 4 n{n + l)(2n +1) 8 n\n + IP]
44
Area = lim 5(n) = 6+ 10 + --4
n — ♦cc 3 3
63. /W = a:2 + 3, 0 < X < 2, « = 4
Let c,. = ' ' '■
65. /W = tan ;c, 0 < ;c < -. n = 4
Letc,
X,. + JC,._
Ax = -, c, = - c
2 /f<^3 4-^4 4
4' ^ 4
4
Area«X/('^,)Ax=2t,' + 3]^
2
69
8
^')-(^-)-(i-)-(^-
.77 77 377 577 777
Ax = — , Ci = — , C2 = TT, c, = T:r, c^
Area - 2/(0 ^=E(tanc,)h^
4
I
1=1 /=1
77/ 77
377
32
—I tan— + tan :7:r + tan XT + tan-:;:^! « 0.345
577
32
777
32
67. /(xj = v^ on [0, 4].
n
4
8
12
16
20
Approximate area
5.3838
5.3523
5.3439
5.3403
5.3384
(Exact value is 16/3)
Section 4.2 Area 187
69./(;c) = tan(^)on[l,3].
n
4
8
12
16
20
Approximate area
2.2223
2.2387
2.2418
2.2430
2.2435
71. We can use the line y = x bounded by x = a and x = b.
The sum of the areas of these inscribed rectangles is the
lower sum.
The sum of the areas of these circumscribed rectangles is the
upper sum.
a b
We can see that the rectangles do not contain all of the area in
the first graph and the rectangles in the second graph cover
more than the area of the region.
The exact value of the area lies between these two sums.
73. (a)
8--
6- ^^^
4- /^^ H
' i >
_ [ }... .If. .i
i ' ■ -
Lower sum:
5(4) = o + 4 + 55+6=15|=f« 15.333
(c)
(b)
Upper sum:
S(A) = 4 + 5| + 6 + 6f = 21I7 = ^ « 21.733
(d) In each case, A.x = A/n. The lower sum uses left end-
points, (i — l)(4//i). The upper sum uses right endpoints,
{i)(A/n). The Midpoint Rule uses midpoints, (/ - ;)(4/n).
Midpoint Rule:
M(4) = 2f + 45 + 5f + 6f = fij - 19.403
(e)
n
A
8
20
100
200
s{n)
15.333
17.368
18.459
18.995
19.06
S{n)
21.733
20.568
19.739
19.251
19.188
M{n)
19.403
19.201
19.137
19.125
19.125
(f) s(n) increases because the lower sum approaches the exact value as n increases. S(n) decreases because the upper sum
approaches the exact value as n increases. Because of the shape of the graph, the lower sum is always smaller than the
exact value, whereas the upper sum is always larger.
188 Chapter 4 Differentiation
75.
77. True. (Theorem 4.2 (2))
I ■! J J 4i >
b. A = 6 square units
79. f(x) = sin X
M
Let A, = area bounded by f(x) = sin x, the x-axis, x = 0 and x = tt/I. Let Aj = area of the rect-
angle bounded by .v = L y = 0, .«: = 0, and x = Tr/2. Thus, Aj = (7r/2)(l) = L570796.
In this program, the computer is generating N^ pairs of random points in the rectangle whose
area is represented by Aj. It is keeping track of how many of these points, N,, lie in the region
whose area is represented by Aj. Since the points are randomly generated, we assume that
N.,
A,
N.
■A,.
1
fW =
sm(xy^ (f'l)
0.75-
/^/, :,'
0.5-
/
/./,:/ "
025-
/
'
Z. JL
The larger Afj is the better the approximation to A,.
81. Suppose there are n rows in the figure. The stars on the left total 1 + 2 + •
n{n + I) stars in total, hence
+ «, as do the stars on the right. There are
2[1 + 2 + •
1 + 2 +
+ «] = n{n + 1)
• + n = \(ri){n + I).
83. (a) y = (-4.09 x \Q-'^)x' + 0.016.x2 _ 2.67x + 452.9
(c) Using the integration capability of a graphing utility,
you obtain
A « 76,897.5 ft^.
(b)
Section 4.3 Riemann Sums and Definite Integrals
-5 .->
1. f{x) = ^x.y = 0,a: = 0,x = 3, c, = ^
n^
3^ _ 3(i - 1)^ _ 3 ,
lim J;/(c,)Ax, = lim % JK\{2i " D
n ->oc /s'l 1 ->oc ,^| v n~ n^
= lim ^2(2'^-')
= lim
3V3
n{n + l)(2n + 1) n{n + 1)
]
-i'^j^K^^'^^-'-m
= 373
;i-o].
2v^ = 3.464
I '„ I.. ' ..,1 > '
3 3(2)' 3(>.-l)^^
Section 4.3 Riemann Sums and Definite Integrals 189
3. V = 6 on [4, 10]. (Note: Ax = — - — = -. ||A|| -* 0 as n ^ oo
n n
|/w--|/(-!)(M<M:?-'
f
6dx= lim 36 = 36
5. y = jc3on[-l, 1]. ( Note: A.T = —^ = - ||A||^0 as h-»oo
n^^, n^ .4', n^ .4',
, 6/ 12/- 8/3"
■1 + ~ + ^
n n- n'
+ 6(l+iU4(2+^ + AU4fl+^ + ^^ 2
n n^ n
r
)^ dx = lim - = 0
7. >' = x2 + 1 on [1,2]. Note: At
2-1 1
>0 as n—^oo
2/(c,)A^. = i:/ 1 +
/\/i
1 + -I + 1
n
, 2/ /2 ,
1 + — + — + 1
n n-
n-,-^, «^e'l \ "/ 6\ /! n-l 3 2/! 6??-
I
10 3 1
^^'^'^'^ = }T^[-j^2;t^^-)= ,
10
9. lim y (3c,. + 10) At, = (3;c + 10) dx
on the interval [— 1, S].
11. lim y 7c,- + 4 At, = Jx^ + 4
ti!-.o ,-e,
-dr
on the interval [0, 3].
13. -idx
Jo
-f
(4 - Lxl) dx
17. (4 - r-) dx
I
19. smxdx
Jo
v^ dy
23. Rectangle
A = W; = 3(4)
Jo
Adx = 12
^^vsxx -.N sN ^^ Rectangle
^^f^
190 Chapter 4 Integration
25. Triangle
A = ^bh = ^4)(4)
-i
xdx=%
27. Trapezoid
A =
b, + b. /5 + 9
r
^h
{2x + 5)dx= 14
^4M
''yj&i Trapezoid
^
^ — h
1 2 3
29. Triangle
A = ^M = ^(2)(1)
/:
(1 - \x\)dx= 1
31. Semicircle
A = ^■TTr^ = |77<3)2
£
V9^^A = ^
4- - Semicircle
r4 ri /•4
In Exercises 33-39, x^dx = 60, Lcrfjc = 6, dx = 2
jfiix = —6
33. L<:dc= -
37. U - 8) iic = ;c ate - 8 d:^ = 6
8(2) = - 10
41. (a) /U) dx = fix) dx+ /(;c) etc = 10 + 3 = 13
(b) I /(;c)dx= -j /(;c)<ic= -10 •
(c) \f{x)dx = Q
(d) 3/(x) otc = 3 /(jc) (it = 3(10) = 30
Jo Jo
35,
39.
.[4.^ = 4/;
xdx = A{6) = 24
I (ix3 - 3x + 2J^ = I x^dc - 3 xdr + 2 dx
= -(60) - 3(6) + 2(2) = 16
43. (a) I [fix) + gix)'\ dx= \fix)dx+ \ gix) dx
= 10 + (-2) = 8
(b) J [gix) - fix)] dx=\ gix) dx - yix) dx
= -2 - 10 = -12
J Igix) dx = 2\
(c) Igix) dx = 2\ gix)dx=^ 2(-2) = -4
(d)
I 3/(x) dx^sl fix)dx = 3(10) = 30
45. (a) Quarter circle below X-axis: — jitt^ = —-^tKX)^ = —tt
(b) Triangle: ^_bh = 5(4)(2) = 4
(c) Triangle + Semicircle below x-axis: -2(2)(1) - 5^2)^ = -(1 + 27r)
(d) Sum of parts (b) and (c): 4 - (1 + 27r) = 3 - 27r
(e) Sum of absolute values of (b) and (c): 4 + (1 + 27r) = 5 + 2ir
(f) Answer to (d) plus 2(10) = 20: (3 - 2ir) + 20 = 23 - 2ir
47, The left endpoint approximation will be greater than the
actual area: >
49. Because the curve is concave upward, the midpoint
approximation will be less than the actual area: <
Section 4.3 Riemann Sums and Definite Integrals 191
51. /W
x-4
is not integrable on the interval [3, 5] and /has a
discontinuity at ;c = 4.
^— :^ — ^'
a. A = 5 square units
55. y
r-t-T'
Jo
d. I IsmiTxdx'- 2^1)(2) « 1
I
57. xjl - xdx
n
4
8
12
16
20
L{n)
3.6830
3.9956
4.0707
4.1016
4.1177
M(n)
4.3082
4.2076
4.1838
4.1740
4.1690
R{n)
3.6830
3.9956
4.0707
4.1016
4.1177
59.
Jo
sin^ X dx
n
4
8
12
16
20
L(n)
0.5890
0.6872
0.7199
0.7363
0.7461
M{n)
0.7854
0.7854
0.7854
0.7854
0.7854
R{n)
0.9817
0.8836
0.8508
0.8345
0.8247
61. True
63. True
67. fix) = ,v' + 3,T, [0, 8]
j^o = 0, ,i:, = 1, X2 = 3, -Xj = 7, .r4 = 8
Ax, = 1, Ar, = 2. Axj = 4, zLv4 = 1
c, = 1, Cj = 2, Cj = 5, c^ = 8
X/(c,) A.t = /(I) A.V, + /(2) Axj + /(5) ^x, + /(8) !^x^
1 = 1
= (4)(1) + (10)(2) + (40)(4) + (88)(1) = 272
65. False
\ (-x)±x =
Jo
192 Chapter 4 Differentiation
69. f(x)
1, jc is rational
0, X is irrational
is not integrable on the interval [0, l]. As ||A|| — > 0, /(c,) = 1 or/(c,) = 0 in each subinterval since there
are an infinite number of both rational and irrational numbers in any interval, no matter how small.
71. Let/(x) = x~,0 < X < 1, and Ajc, = 1/n. The appropriate Riemann Sum is
n n / : \ 2 1 in
1 r,-, r,^ 07 -,1 ,• 1 n{2n + l)(n + 1)
n->co rj' n->oc n' 6
,. 2«2 + 3w + l ,. /I 1 1
= hm T-^ = hm {- + — + —2
)4
Section 4.4 The Fundamental Theorem of Calculus
1. fix) =
3. fix) = xjx~ + 1
xVjc^ + 1 cic = 0
J
1-0=1
f,
7. U - 2) (ic =
>(i-)-(4-
1 \ -
11. ilt- \Ydt= (4f--4f
Jo Jo
Ix
« /I
+ 1) rf/
rr^ - 2f2 + f
1°
3
4-2+1=1
0 3 3
" '-l-aUM-D-l
"■/>'
2 rf/ =
l-"-^ir(l
,3/2
+ 2 I = -4
' = if 1 - 2
0 3\2 3
J_
18
21. I (r'/3-f2/3)rf,=
1,4/3 _ lf5/3
4 5
]>»
|2j:- 3| A:= (3 - 2x) d:x + (:
Jo Jo J3/2
3 3\ _27
4 5) 20
23. |2j: - 3| (ir = I (3 - 2x) lic + (2j; - 3) a!^; split up the integral at the zero x =
= [
3x- x^
3/2
_3.];^.f?-2)-0 + (9-9)-f^-?U2f|-^Ul
3/2 \2 4
4 2/ \2 4/ 2
Section 4.4 The Fundamental Theorem of Calculus 193
25. I |.r2 - 4| <ir = I (4 - x'-) dx + I (.t^ - 4) dx
0
-I
f-1
-|8-;U(9-12)- ^3
23
3
Jo
27. (1 + sinx}dx =
X — cos X
(tT + 1) - (0 - 1) = 2 + TT
29.
-77/6
sec- .t dx =
.-77/6
tan.x
77/6
-,7/6
73 / v^\ 273
r'7/3 r T^/^
31.
4 sec 6 tan 0
. -77/3
de =
4 sec e
= 4(2) - 4(2) = 0
-V3
10.
Jo
33. 10.000(r - 6)di = 10,000
- 6t
5135,000 35. A = (.v - x-) dx
' ^1
0 6
37. A = (3 - .v)7^a[t = (3.t'/2 - .rV2) j^. =
Jo Jo
2^-3/2 _ 2,.5/2
-(10 - 2v)
1273
I
77/2
39. A = I cos X dx =
sin.x
77/2
41. Since y > 0 on [0. 2],
A = {3x- + 1) dx =
= 8 + 2 = 10.
43. Since y > 0 on [0, 2],
A = (.t^ + x) dx
Jo
x" x'-
4^2
4 + 2 = 6.
45. (x - 27^)rf.v =
X^ 4^3/2-
/(c)(2 - 0) =
c - 27c =
.2 3
6 - 8^2
3
3 - 4.-2
872
c-2V^+ 1 = - .^" - + 1
{Vc-l
,2 6 - 4^'2
./^-1 =
/6 - 4x 2
*V^]
c« 0.4380 ore = 1.7908
194 Chapter 4 Integration
Tr/4
47. I 2 sec^ xdx =
J—TTj
■/4
fic]
J-7r/4
2tanx = 2(1) - 2(-l) = 4
f)]
2 sec- c = —
2 4
sec c = ±-
c = ±arcsec(— ^
= ±arccos -— « ±0.4817
49.
1
2 - (-2)
Average value
£'"-"
)& =
4;t - -j:^
273
4 — jc- = r when jc- = 4--orx = + ^
3 3 3
+ 1.155.
sin x ate
Average value = —
TT
sinx = —
77
= 1
0 77
(-¥•!) 3 (^-f)
53. If/ is continuous on [a, b] and F'U) = f{x) on [a, fc],
(0.690, 1)
then fix) dx = F(b) - F{a).
x« 0.690, 2.451
Jo
55. I f{x)dx = - (area of region A) = -1.5
Jo
/•6 (-6
59. I [2 + /(jc)] dx= 2dx+ \ f(x)dx
= 12 + 3.5 = 15.5
61. (a) F{x) = ksec-^x
F(0) = A: = 500
Fix) = 500sec2x
57.
f \fix)\ dx = -Ifix) dx + Ifix) dx =
Jo Jo Jl
(b)
1^/3 -oJo
500 sec- xdx =
1500r l'^/^
1500,
[tanx]^^'
(v^-O)
77
= 826.99 newtons
~ 827 newtons
63.
-oJo
(0.1729r + 0. 1552^2 - 0.0374^3) dt '^ ^
0.08645r2 + 0.05073?^ - 0.00935?^
0.5318 liter
Section 4.4 The Fundamental Theorem of Calculus 195
65. (a)
The area above the j:-axis equals the area below the
jf-axis. Thus, the average value is zero.
The average value of S appears to be g.
67. (a) V = -8.61 x 10"^ + 0.0782^ - 0.208r + 0.0952
(b) ^
(c)
69. F(x)
Jo
-8.61 X lO-V 0.0782r3 0.208^2
+ 0.0952f
5r
0 2
F(2)=--5(2)= -8
f(5)=^-5(5)=-^
F(8)=y-5(8)=-8
73. F{.x) = I
cos ddd = sin 6
sin X - sin 1
F{2) = sin 2 - sin 1 = 0.0678
F(5) = sin 5 - sin 1 -= - 1.8004
f(8) = sin 8 - sinl =0.1479
= 2476 meters
71.FW = 0^v = £lOv-.V=^];
i°+ 10=10 1-1
X \ X
F{2) = 10(^-j = 5
f(5) = 10^-j = 8
F(8) = 101
7\ 35
75. (a) \ {t + 2)dt= [l
+ 2t
= —X- + Zx
77. (a)
3
-,4/3
4
' = |(.^4/3 _ 16) = 3 ,/3 _ j2
8 4 4
(b) 4f^/3 - 12
= rl/3 = 3/
79. (a) sec^ r A = tan r
J.r/4 L Ax'A
tan .t — 1
(b) — [tan.t - IJ = sec-.r
81. F(x) =1 (f- - 2t) dt
F'(x) = x^ -2x
83.
. F{x) = Vr» + 1 df
F'(.v) = Vx^+ 1
= 1'
85. FU) = I t cos tdt
F'{x) = .TCOSX
196 Chapter 4 Integration
87. Fix) = {4t + \)dt
I
= [2(.r + 2)2 + (x + 2)] - [2x2 + x]
= 8.V + 10
F'(x) = 8
rsi
89. f U) =
Jo
Jtdt
23/2
3
= f(sinx)V2
0 3
F'W = (sin j:)'/2cosx = cos-tv/sinx
Alternate solution
Fix)
Jo
^tdt
F'ix) = Vsin JT -p (sin x) = Vsin j:(cos x)
ax
Jo
93. g{x)= \fit)dt
giO) = 0, gH) = ^, g(2) - 1, gi3) » ^, g(4) = 0
Alternate solution:
Fix) = (4f+ l)rff
ro rx
= (4r + 1) rff +
(4f+ 1)A +
0 Jo
(4r + \)dt+ i4t + 1) rff
(4f+ 1)A + (4f + 1)A
I Jo
F'ix) = -i4x + 1) + 4(x + 2) + 1 = 8
Jo
91. Fix) = I sin f 2 rff
Jo
F'ix) = sin(x3)2 • 3^2 = 3x^ sin x^
95. (a) Cix) = 50001 25 + 3 | f''"' dt
[4
3(25 . 3/;,
= 5000 25 + 3
Ij5/4
= 5000(25 + y^'") = 1000(125 + 12x5/")
(b) C(l) = 1000(125 + 12(1)) = $137,000
C(5) = 1000(125 + 12(5)5/4) ^ $214,721
C(10) = 1000(125 + 12(10)5/") ^ $338,394
g has a relative maximum at x = 2.
97. True
99. False
; I x-^dx = J x-^dx + I x-2atc
Each of these integrals is infinite. /(x) = x ^
has a nonremovable discontinuity at x = 0.
By the Second Fundamental Theorem of Calculus, we have
1
f'ix) =
lU '
(l/x)2 + 1\ xV X2 + 1
' 4-^^ = 0.
1 + X2 x2 + 1
Since fix) = 0,/(x) must be constant.
Section 4.5 Integration by Substitution 197
103. x{t) = t^ - 6t^ + 9t - 2
x'{t) = 3f2 - I2t + 9
= 3(f2 - 4/ + 3)
= 3(r - 3)(t - 1)
Total distance = |j:'(')|rf'
Jo
= \'3\{t-3){t-l)\dt
Jo
= 3 J (i- - 4t + 3)dt - 3 (f2 - 4r + 3)A + 3 (t^ - 4t + 3)dr
= 4 + 4 + 20
= 28 units
105. Total distance = \x'{t)\dt
= I Ht)\dt
= 2(2 - 1) = 2 units
Section 4.5 Integration by Substitution
f{g{x))g'{x)dx
u =g{x) du = g'{x)dx
1. I {5x^ + \Y{\Qx) dx 5x- + \
x'+ 1
I tan- X sec- x
dx
tanjc
XQx dx
2x dx
sec^ X dx
/'
7. 1(1 +2.t)-'2d:t= ^' \^^' + C
Check: ^
(1
^ + C] = 2(1 +
ixy
». I (9 - x2)'/2(-lv) dx = ^^ , J'^' + c = ^(9 - .v-)3/- + C
Check: -^[|(9 - .r^)^/: + c
dxi3
3/2 ■ ^ 3
2 3
3 2
(9_^)i/:(_2^)= y9-.r=(-lT)
198 Chapter 4 Integration
11. f.r3(.x^ + 3P<ic = ^f(
.r3(.x^ + 3Ydx^^\(x' + 3)2(4x3) ir = - ^"^ t ^^' + C = ^"^ .t ^^' + C
4 3
12
Check:
£[^^ + c]=^^^(4x3) = (^ + 3)V)
13. |.rV-l)^A = |J(.
.r2(x3 - 1)" A = :^ I (x^ - Ij-'CSx^) dx = ^
w-m^c = ^^^^^^c
-]
15
Check:
^
Cv^ - 1)
15
+ C
5(.r3 - 1)^(3x2)
15
= x2(x3 - I)'*
rVFT2 dt = ^ I (r^ + lY'^lt) dt = \^^^^ + C= ^^±^ + c
15. I rVFT2 * = I r
"(r2 + 2)3^
Check:
dt
+ C
3/2(f2 + 2)'/'(2f)
(?2 + 2)'/2t
17. [5x(l-x2)'/3d^c=-|J(l-x^;
Check: 4[-^l - x^)''/^ + c
5 (1 _ y2W3 15
)>/3(-2x) ^ = -^ • ^ ^^3^ + C = -yd - x2)V3 + c
15 4,
--^ • ^1 - x2)i/3(-2x) = 5x(l - x2)i/3 = 5xyr^
O 3
19- J(]-f]^^ = -||(1 - x')-\-2x)dx= -
1 (1 - x^
+ C^
4(1 - x-:
+ C
Check:
^
.4(1 - x^Y
^+ c
= i(-2)(l-xr'(-2x) = ^^-:^
-k
1
+ ^3)2<i^ = jjd + ^')-'(3^-) 0^^ =
(1 + x^.
-1
+ c
1
3(1 + x3)
+ C
Check:
dx
1
L 3(1 + x3)
+ C
-i(-i)(i+x3)-w) = ^Y:^
23.
/^%-=-i/<'-
,<i,--jJ(l-x")-'»(-2«)i<.-iiL_i!
2-1 1/2
+ C= - Vl - x2 + C
Check: —[-(1 - x^)'/^ + C] = --(1 - x2)-'/2(-2x) = -y^==
dx 2 yn^
25.
[1 + (lA)?
[1 + iUt)V
4
Check:
dr
+ C
+ C
l.-J. . IV/ 1\ 1/. , 03
4(4)i+7jl-^j = ?l'^7
(2x)
^^•/;fc^=iK"^^4[
Check: ^V2i + C] = ^(2x)-'/2(2)
1/2-
1/2 J
+ C= V2x + C
dj:'-
vs
Section 4.5 Integration by Substitution 199
29.
: ^W^r- + 2;c3/2 + 14.'/= + cl = :n±^±I
+ 35) + C
Check;
31.
j r^ff - 7) * = f ('' - 2;) * = |f4 - r- + C
Check:
dt
-t" - t^ + C
= f3 - 2f = f-U - y
33. [(9 - y)^dy = [(Qy'/^ - y/z) rfy = 9^|y3/2
|v5/2 + c = |v3/2(15 - >') + C
Check: — 7^/^(15 - y) + c
6^/2 _ 1^/2 + c
^^•^=/h^7ife]
tic
v/l6~- x^
= 4lxdx- lliie- x-)-^'H-2x)dx
-(f)-
(16 - x2)i/2
1/2
+ C
2x2 _ 4716 - X- + C
= 9v'/2 - y/2 = (9 - y)v^
^^•^ = J(.v2 + lx-3P^
- + 2.x - 3)-2(2x + 2) dx
{x- + 2.x - 3)-
2(.x2 + 2x - 3)
+ C
c
39. (a)
dv
(b) ^ = .xV4^^, (2, 2)
dx
-/■
^/'
xv^4^^dx = -^ (4 - .x2)i''-(-2xdx)
= -i • ^(4 - .r2)V2 + c = -{(4 - x-)^'- + c
(2, 2): 2 = -^(4 - l-y- + C => C = 2
V = -^(4 - .x2)3/2 + 2
41. J 77 si
sin TTxdx = — cos ttx + C
45.j^_coslde=-jcos\[-j,)de
-sin- + C
6
43.
I sin Iv ix = - I
(sin 2jc)(2;t) dx = -- cos Ix + C
200 Chapter 4 Integration
47. \smlxcos.2xdx-=^\ (sin 2x)(2 cos 2x) dx = - i^iH^L + C = - sin^ 2x + C OR
\^^^ + C, = -\cos^lx + C, OR
sin 2x cos 2xdx = rr I 2 sin 2x cos Ixdx = ir I sin 4j: A — - cos 4j: + C-,
49. I tan"* .v sec- xdx = ^^ + C = - tan^jc + C
. I sin 2x cos 2j; A; = — I (si
I sin 2x cos 2x dx = — - I (cos 2j:)(-2 sin 2x) dx = — -
/si.2.cos2,A = i|2™2xc.s2.<*:.i|:
I
I CSC^ X I
51. — —dx = - (cot j:)"3(-csc2.r) otc
= -i£5L^ + c = —^ + C = )-tm^x + C = i(sec2;c - 1) + C = |sec2;c + C,
— 2 2 cot^ X 2 2 2
53. cot' xdx = \ (esc- j: - 1) A = - cot .r - jc + C 55. /(x) = cos ^ (it = 2 sin ^ + C
Since /(O) = 3 = 2 sin 0 + C, C = 3. Thus,
/(;c) = 2sin| + 3.
51. u = X + 2, X = u ~ 2,dx = du
\xjx + 2dx = \(u - 2)^du
/"
3/2 _ 2„l/2) J„
= |«V2 _ l„3/2 + c
= ^(3« - 10) + C
= ^x + 2)3/2[3(x + 2) - 10] + C
= ^jc + 2)3/2(3;c - 4) + C
59. u = 1 — X, X = I — u, dx = — du
I du
jjcVl -Xtic = - 1(1 - Mfv^
= - J(«>/2 -
2m3/2 + „5/2) ^j,
= -(|m3/2-1«V2 + |„7/2) + c
2«3/2^
105
■(35 - 42m + 15m2) + C
-r^(\ - xy/^[35 - 42(1 - x) + 15(1 - x)^] + C
-j|^(l - x)3/2(15a:- + 12x + 8) + C
Section 4.5 Integration by Substitution 201
61. M = 2x- l,x = r{« + l),dx = -zdu
J J2x - 1 J V" 2
!<3/2 + 2ttl/2 _ 3„-l/2) ^„
= |(|m5/2 + |„3/2 _ 6„1/2J + c
1/2
= ^(3m= + 10« - 45) + C
60
V2x- 1
60
{?>(2x - 1)' + \Q{2x - 1) - 45] + C
-^Jlx - \{\2x- + at - 52) + C
60
^V2jc - 1(3^2 + 2;^ _ 13) + c
63. « = a: + 1, .X = M — I. dx = du
/
: oLv = ;=^ du
(X + 1) - V.V + 1 J M - V«
(y;i + 1)(7^- 1)
7ii(v«- l)
(1 +«-i/2)rfu
rf«
= -(« + 2«'/2) + C
= -M - 2V« + C
= -(.r + 1) - 2V-r + 1 + C
= -.r - iVxTl - 1 + C
= -(.v + iJTTl) + c,
where Cj = — 1 + C
65. Let u = x~ + I, du = 2v dx.
J x(.r2 + 1)3 dr = ^1 (;c2 + m2x) dx
U- + 1)^
67. Let H = .v' + 1, rf" = 3.r- dx
Ix-J^P^Tldx = 2 • I Cr' + l)»/-(3.r2)a^t
Ls 3/2 J,
= ^[27-2./2]=12-§V2
202 Chapter 4 Integration
69. Let M = 2jc + 1, Jm = 2 dx.
r4
Jo ji^n 2jo
^ dx = ]-\ {lx+ \)-"\2)dx =
Jl.
;x + i[ =V9-VT = 2
71. Let tt = 1 + v^, du = — -p dx.
Ijx
73. H = 2 — j:, j: = 2 — M, (ic= —du
Whenx = \,u= \. Whenx = 2, w = 0.
1
2v^/^=rTT7^Ji-~2+'-2
['(;C - 1)72"^ A = [" - [(2 - m) - iJv^dM = f («3/2 - „l/2) ^„ = |,,5/2 _ |„3/2l° = _ r| _ |1 =
_4_
15
I
r/2
75. I cosl-jf)(ic
3 . 1
— sin -X
2 V3
V2
3/73\ 3v^
77. u=x+\,x = u— \,dx = du
When .t = 0, M = \. When x = 7, m = 8.
Area
^M
xi/x + ldx= {u - \)l/u
= I {u"'^ - m'/^) du
-uy^ - -u^l^
384
12
3 _ 3\ 1209
7 4/ ~ 28
79. A = I (2 sin jc + sin 2;c) ate = - 1 2 cos * + r cos 2a;
'. A = I (2 sin jc + sin It) ate = — 2
^£:Mf)--£>(f)©-[--(c:-(^-')
81. Area
83.
Jo
Jc , , ,,, 10
, dx^ 3.333 = —
Jlx + 1 3
r
85. xjx - 3dx'= 28.8
144
f
Jo
87. I |0 + cos^lrfe = 7.377
r^
J
89. \{2x- \y-dx = ^\{2x- \yidx = h2x - 1)^ + C, = |x3 - 2^:^ + .r - ^ + C,
». j(2x- \r~dx = ]A(
\{2x- \)-dx = (4;c--4x+ 1)
They differ by a constant: Cj = C,
i& = —x^ — 2jc + X + C2
1
6'
Section 4.5 Integration by Substitution 203
91. fix) = x\x^ + 1) is even.
x\x^ + 1) ir = 2 [x" + x'-)dx = 2
5 "^ 3
32 8
T + 3
272
15
93. fix) = x{x'^ + 1)3 is odd.
xix- + 1)3 iv = 0
Jo
95. x'±x
Jo
(a)
(c)
the function x^ is an even function.
(b) I x2iT = 2j>dLr = y
(d)
3^:^ ^- = 3 x'-
dx= 8
97.
U' + 6x- -2x-2)dx= ix^ - 2x) dx + (6.r- - 3) rf.t = 0 + 2 i6x- -
3)dx = 2
2x' - 3.x
232
99. Answers will vary. See "Guidelines for Making a Change of Variables" on page 292.
101. fix) = xix- + 1)2 is odd. Hence
, j^xix^ + D-
dx = 0.
dV
103. ^
dt it + 1)2
Vit
( k , k ^
V(0) = -A: + C = 500,000
V'(l) = -i't + C = 400,000
Solving this system yields A' = —200.000 and
C = 300,000. Thus,
V(r) = ^ + 300.000.
When t = 4, V'(4) = $340,000.
105.
b - a
f
74.50 + 43.75 sin ^
6
dt =
b - a
^^ ^„ 262.5 tTt
74.50/ cos — -
77 6
(a)-
(b)
,, ,„ 262.5 Trt
74.50/ cos —
77 6
262.5 T7T
74.50/ cos —
77 6
3 1/ ''62 5 \
= - 223.5 + - — - I = 102.352 thousand units
0 3\ 77
^'447 + ^^^ - 223.5 I = 102.352 thousand units
77
'''T2
262.5 777
74.50/ cos —
77 6
'- 1 / 262.5 262. 5\
= — 894 + 1 = 74.5 thousand units
0 12\ 77 77
204 Chapter 4 Integration
b -
(a)
(b)
(c)
-S:
1 c 1 r 1 1 1*
107. I [2 sm(607rf) + cos(1207r;) ] dt = -ttt- cos(60iTf) + -— — sin(120Trt)
b — a\_ jOtt 12U TT
^^
1
(1/60) - 0
1
-^cos(60..) + Tisi„(12O.0]7 = 60[(^ + o) - [~^^] = ^ » 1.273 amps
(1/240) - 0
1 1 i'/^'"' r
cos(6077-f) + -— - sin(120-7rf) = 240
12077 Jo L
3077
' + '
30V2ir 12077/ V 3077,
-(5 - 272) ^ 1.382 amps
1
(1/30) - OL 3077
-^7— cos(6077t) +
I2O77
sin(12077.)];^° - '{[£) - (-i)] - 0
I amps
109. False
|(2x+ \f-dx = ]A{2x+ \f2dx = hlx+ If + C
111. True
rio rio rio
(ax' + bx- + ex + d) dx = \ {ax^ + ex) dc + {bx^ + d) dx = Q + 1
J-10 J-10 J-10
Odd Even
J-10
(to- + d)dx
0
113. True
4 sin a: cos xdx = l\%\n2xdx = — cos 2jc + C
4- sin a: cos j: lic = 2 [ s
115. Let « = j: + /i, then du = <ic. When j: = a, « = a + ft. When x = b, u = b + h. Thus,
J-fc rb + h rb + h
f(x + h)dx=- f{u) du = fix) dx.
a Ja + h Ja + h
Section 4.6 Numerical Integration
;.6667
I. Exact: JJ,. ^ = [i^]^ | » 2.,
Trapezoidal: j x^ ^ - ^[o + 2^)' + 2(1)= + ll^J + (2)A = ~- = 2.7500
Simpson's: J jc^otc - ifo + 4^^)^ + 2(1)^ + 4(|j' + (2)^1 = | - 2.6667
3. Exact:
Trapezoidal
Simpson's
4.000
+ 2(^)\ 2(1)3 + 2(1)' + (2)3
17
4.2500
x^dx^
0 + 4(|y + 2(1)3 + 4(1]' + (2)3
^ = 4.0000
Section 4.6 Numerical Integration 205
5. Exact: \ }? dx = \KA = 4.0000
Trapezoidal: I x' oLr ~ -
o-iy-(!r-er-»)-er-(!r-er^
Si™p».V [^ * . ^[O + 4(i)' . 2g)' . 4g)' . 2(,). + 4(5)' + 2(f)' + 4(^)' + 8
4.0625
4.0000
7. Exact:
/:
-Jxdx
:,j/2
^18_ 1^ = ^.12.6667
4 3 3
Trapezoidal: I ^dx =
Simpson's: I Vx dx
,37 ^ /2 ^ /47 ^ /26 ^ /57 ^ /31 ^ /67 ,
4 V8 \'4 VS V4
- 12.6640
2 + 4^/? + V2T + 4,/^ + V26 + 4^/" + VJT + 4^/? + 3
= 12.6667
9. Exact:
1U +
;(ix
X + 1
- Ill
= -T + :;■ = 7 = 0.1667
1 3 2 6
Trapezoidal
Simpson's:
i
(x + 1
' ^«^
' . 2f.....l-^U 2( '
4 ^((5/4) + \r-j A((3/2) + l)-y n((7/4)
•) + 1)-] ^ 9J
KMf-l^f^^--
J, U + 1)=
ix
^.41
1
+ 2
1
4 "V((5/4) + If) -V((3/2) + 1)V \((7/4) + 1)^
+ 41
1
p)4]
^i^lMi^i^f^i)--
11. Trapezoidal: Vl + a^ dx - ^[l + 2^1 + (1/8) + ijl + 2j\ + (27/8) + 3] = 3.283
Jo 4
f
Simpson's: | Vl + x" dx « -[1 + 4Vl + (1/8) + ijl + AJ\ + (27/8) + 3] = 3.240
Graphing utility: 3.241
13. JxJ\ - xdx= Jxi\ - x) dx
Jo Jo
Trapezoidal
Jo
x) dx ~ —
I
Simpson's: I ^x(l — x) dx ~
0+4^
■-!
i._i„, ,.
vf4)-VlH)
« 0.342
= 0.372
Graphing utility: 0.393
206 Chapter 4 Integration
15. Trapezoidal:
/:
t/2 J'ttFI
cos(x^) dx = • — - —
== 0.957
cos 0 + 2 cosi " . I +2 cos
^V2V , ^ ..Y3VV2V
+ 2 cosI
+ cos .. / —
r
Simpson's: | cos(a:^) dx = — rr —
== 0.978
cos 0 + 4 cosI " / 1 + 2cos( r | + 4 cost V—\ + cosI
Graphing utility: 0.977
\:
Yl. Trapezoidal: sin jc^ ^ ^ -sTrCsinll) + 2 sin(1.025)2 + 2 sin(1.05)2 + 2 sin(1.075P + sin(l.l)2] -= 0.089
80
r
Simpson's: | ^mx^dx'^ Tio'-^'"^^^ "^ 4sin(1.025)2 + 2 sin(1.05)2 + 4sin(1.075)2 + sin(l.l)2] == 0.089
Graphing utility: 0.089
19. Trapezoidal:
/:
r/4
X tan X dx ■
32
0 + 21
16
^^"(5)^2(7!)^
^U2f^ltani
16
'16
f)n]--
194
Simpson's: \^\ tanxdx^ ^[o + 4(^) tan(:^) + 2(ff ) tan(ff ) + 4(^] tan(ff ) + j] » 0.186
Graphing utility: 0.186
21. (a)
The Trapezoidal Rule overestimates the area if the graph
of the integrand is concave up.
23. fix) = x^
fix) = 3x^
fix) = 6x
fix) = 6
f'Kx) = 0
(2 - 0)3
(a) Trapezoidal: Error < ,^. (12) = 0.5 since
/"W is maximum in [0, 2] when jc = 2.
(2 - 0)^
(b) Simpson's: Error
/^W = 0.
180(4")
{0) = 0 since
25./"(x) = ^in[l,3].
(a) \f"{x)\ is maximum whence = 1 and |/"(1)| = 2.
2'
Trapezoidal: Error < 7^(2) < 0.00001, n^ > 133,333.33, n > 365.15; let« = 366.
/W'(;c)=^in[l,3]
(b) \p'*>{x)\ is maximum when jc = 1 and [/"'(l)! = 24.
2'
Simpson's: Error < 75777(24) < 0.00001, «■* > 426,666.67, « > 25.56; letn = 26.
Section 4.6 Numerical Integration IQTI
27. f(x) = ym
(a) f"(x)
1
4(1 + x)V2
in [0, 2].
\f"{x)\ is maximum when.t = 0 and |/'(0)| = -.
Trapezoidal; Error < Ty^lj) < 0.00001, n- > 16,666.67, « > 129.10; let « = 130.
(b) /(*>W =
-15
16(1 + xy/^
in [0, 2]
|/('*>(jc)| is maximum when a: = 0 and \f'^\Qi)\ =
16'
Simpson's: Error < 7^77(77 I < 0.00001,/?'' > 16,666.67, n > 11.36; let « = 12.
I0O/2 \16
29. fix) = tan(jc2)
(a) f"(x) = 2 stc\x-)[\ + Ax- tan(x2)] in [0, 1].
\f"(x)\ is maximum when .r = 1 and |/"(1)| = 49.5305.
Trapezoidal: Error < .7 ^ (49.5305) < 0.00001, n- > 412.754.17. n > 642.46; let n = 643.
(b) /'"'(x) = 8 sec2(jt2)[12T:2 + (3 + 32^) tan(x2) + 36x- tan2(.:c2) + A^x^" tan^i^?)] in [0, 1]
\f''\x)\ is maximum whenx = 1 and |/'"(1)| " 9184.4734.
0-0)5,
Simpson's: Error <
180W'
-(9184.4734) < 0.00001, «'' > 5,102,485.22, n > 47.53: let n = 48.
31. Let/(x) = Ar* + Bx- + Cx + D. Then/'^H^) = 0.
Simpson's: Error < . (0) = 0
180/2
Therefore, Simpson's Rule is exact when approximating the integral of a cubic polynomial.
1
Example:
I
x^dx =
r /iv 1
1
0+4^ +1
L 12^ J
4
This is the exact value of the integral.
33. f(x) = 72 + 3.t- on [0. 4].
n
LM
M{n)
R(n)
Jin)
Sin)
4
12.7771
15.3965
18.4340
15.6055
15.4845
8
14.0868
15.4480
16.9152
15.5010
15.4662
10
14.3569
15.4544
16.6197
15.4883
15.4658
12
14.5386
15.4578
16.4242
15.4814
15.4657
16
14.7674
15.4613
16.1816
15.4745
15.4657
20
14.9056
15.4628
16.0370
15.4713
15.4657
208 Chapter 4 Integration
35. fix) = sinv^on[0,4].
n
Lin)
Min)
Rin)
Tin)
Sin)
4
2.8163
3.5456
3.7256
3.2709
3.3996
8
3.1809
3.5053
3.6356
3.4083
3.4541
10
3.2478
3.4990
3.6115
3.4296
3.4624
12
3.2909
3.4952
3.5940
3.4425
3.4674
16
3.3431
3.4910
3.5704
3.4568
3.4730
20
3.3734
3.4888
3.5552
3.4643
3.4759
37. A
Jo
ix COS X dx
Simpson's Rule: « = 14
r,r/2
^ V,cos.:.fa==-|^70cos0 + 4^-cos- + 2^-cos- + 4^-cos-
0.701
' TT ■7T\
TT cos —
2 2
Jo
39. W= 100xVl25 - x^dx
Simpson's Rule: « = 12
100xVl25 -x^dx'^ -^
Jo 3(12)
-«'^)v--(^r-ov— (f
'^^(u)-\^~^ ~ [nf + • • • + O] - 10,233.58 ft • lb
41.
-0
:dx Simpson's Rule, n = 6
■Jr "^ ^,^. [6 + 4(6.0209) + 2(6.0851) + 4(6.1968) + 2(6.3640) + 4(6.6002) + 6.9282]
= :^[113.098]« 3.1416
36
1000
43. Area - ^(J^^25 + 2(125) + 2(120) + 2(112) + 2(90) + 2(90) + 2(95) + 2(88) + 2(75) + 2(35)] = 89,250 sqm
45. I s\n^dx = 2, « = 10
Jo
By trial and error, we obtain t ~ 2.477.
Review Exercises for Chapter 4 209
Review Exercises for Chapter 4
1. y
./,
3. \{2x'^ + X - \) dx = ^}? + ^rx^ - X + C
/^-/
X H — r;\dx = —x- h C
X-) 2 x
J(4x-
3 sin x) dx = 2x- + 3 cos x + C
9./'U)= -2x, (-1,1)
/W = -Ixdx = -x^ + C
Whenx = -1:
>-= -1 + C= 1
C = 2
y = 2 - ^2
11. a(r) = a
v(f) = a rfr =
ar + C,
v(0) = 0 + Ci = 0 when C, = 0.
v(f) = at
-\'
s{t) = I ar rfr = -r + C;
s(G) = 0 + C, = 0 when C. = 0.
i(f) = 2''
5(30) = -(30)- = 3600 or
. = « = 8ft/sec^.
v(30) = 8(30) = 240 ft/sec
13. a(t) = -32
v(r) = -32f + 96
s(r) = - 16r= + 96f
(a) v(f) = - 32f + 96 = 0 when t = 3 sec.
(b) 5(3) = - 144 + 288 = 144 ft
96 3
(c) v(t) = - 32f + 96 = — when t = - sec.
(d) 5I
(1) =
= -16(|j + 96(|j= 108 ft
15. (a) 2(2'- 1)
(b) S/'
i=l
10
(c) 2(4' + 2)
210 Chapter 4 Integration
17. y = ^r—T, Aat = -, /J = 4
X- + 1 2
S{n) = 5(4) = ^r^ +
10
+ .J^ +
10
2\_ 1 (1/2)2 + 1 (1)2 + 1 (3/2)2 + i_
13.0385
s{n) = s{A)
^h
10
+ -^+ 10
10
_(l/2)2 +1 1 + 1 (3/2)2 +1 22+1
« 9.0385
9.0385 < Area of Region < 13.0385
4
19. y = 6 — jc, Ax = -, right endpoints
Area = lim V f{ci) Ax
lim -
n-*oo n
6n
4 n{n + 1)1
n 2 J
= lim
24-8 ^^-^ I = 24 - 8 = 16
21. y = 5 - x^, Ax = -
n
Area = lim V /(«') Ajc
= lim 2
-2 +■
= lim^2ri+i^-^l
,. 3r 12«(n +1) 9 «(n + 1)(2m + 1)
= hm - n + — ^
n->oo n\_ n 2 n- 6
lim[3 + 18^^-?(^l±il%^^l
n^-oo L n 2 «' J
= 3 + 18 - 9 = 12
23. X = Sy - r, 2 < y < 5, Ay = -
n
Area = lim V
5
(-f)-(-f]
©
= lim - 2
No+*^' 4 12'-
n n
9(^"
= lim - 1
1
->
>
i
6+^-^1
= lim -
^ , 3 n(n + 1) 9 «(« + l)(2n + 1)
"" +
n 2 n 6
18 + --9
27
2
Review Exercises for Chapter 4 211
25. lim y {lei - 3) ^xi = (2x-'i)dx
-f
\(5-\x-5\dx= I (5-{5-x))dx= \ xdx = ^-
(triangle)
29. (a) J [/W + gW] d.x= \ fix) dx + \ g{x) dx = 10 + 3 = 13
(b) f [/W - gW] ^ = J /W dx-\ g{x)dx=\Q-3 = l
I
(c) [2/(;c) - SgW] ^ = 2
J/Wd:x-3[
gW ^ = 2(10) - 3(3) = 11
(d)
5/(^) dx = 5\ f{x)dx = 5(10) = 50
31. j (^ + l)
dx
4
4(16) + 8
]-[l
+ 1
73
(0
Jo
33. {1 + x)dx =
2x +
.^-u
'•/:
35. (4r3 - 2t) dt
r - t-
= 0
37. f ;cv^a^r = f ;t3'2^ = Hr^^'T = Jlv^)' - (74)'] = ^(243 - 32) =
J4 J4 l5 J4 5 5
P'r/4 r
39. sindde = -cos
3 17/4
0
f)+l = l+^ = ^^
2 / 2 2
41. (2v- l)at( = \x
\(x--9)dx= J -9x1'
= (y-36)-(9-27)
64 _ 54 ^ W
3 3 3
212 Chapter 4 Integration
Jo
45. {x- x^)dx =
2 4
' =1-1 = 1
0 2 4 4
47. Area
=r
:dx =
^x L(l/2)Ji
4;c'/2
= 8(3 - 1) = 16
1 f 1 ri p 2 2
49. —pdx = -ijx = t(3 - 2) = - Average value
2 _ 1
5 Vx
25
2 4 6 8 10
51. F'W = ;cVl + ;c3
53. F'W = ;c2 + 3x + 2
f (;c2 + 1)3 die = JC
;c^ 3
55. I (;c2 + \f dx = \{x^ + ■ix^ + l,x^ + \) dx = J + -x^ + x^ + X + C
57. M = a:^ + 3^ ^^ = 3;t:2 ^
I /' att = |(x3 + 3)-'/2;c2^ = tI(-^ + 3)-'/2 3jc2d:^ = |(;c3 + Zf- + C
J Jx" + 3 J 3J 3
59. M = 1 — 3;c2, £/m = — 6j; (&
/
41 - ix-'Ydx= -| (1 - Ix-'Yi-exdx) = -^(1 - 3;c2)5 + C = ^(3;c2 - 1)^ + C
1. /si
61. I sin' X cos xdx = - sin'' jt + C
63. . "" ^ dx = (1 - cose)-'/2sin0de = 2(1 - cos e)'/^ + C = 2^/1 - cos e + C
J J\ - cos 0 J
tan";
tan" ^ ' X
65. I tan" x sec^ xdx ^^ — + C,ni^ -\
n + \
'■/'
i/<
67. [ (1 + sec TTxf- sec Tr;c tan 7rx<36c = — | (1 + sec Trj:)-(7rsec Tr;ctan nx) dx = —(1 + sec ttxY + C
69.
j'xix^ -4)dx = ^j y- 4)(2.) ^ = I^^^Y^]' _ = ^[0 - 9]
Review Exercises for Chapter 4 213
4-2 = 2
71. I , ' dx = I (1 + x)-"^dx= [2(1 + xY'^
Jo VI + ->c Jo L
Ti. u = \ — y, y = \ — u, dy = — du
Whenv = 0, « = 1. Wheny = 1, « = 0.
2tt\ {y + l)Vl -ydy = 2tt\ -[(1 - u) + \]V^du
= 27r| (m3/2 - 2M'/2)d„ = 2T7r|«^''2 - |«
75. [ cos(f ) dx = 2|^ cos(f ) I ^ = [2 sin(f
°_ 28t7
1 15
77. u = \ — x,x = 1 — «, a[r= —du
When x = a, M = 1 - a. When x = fe, « = 1 — /).
Pa,b= \ ^x^/V^~xdx = —\ -i\-u)Judu
3/2
l-a 4
2„3/2
15
(3m - 5)
i-fc
l-a
(1 - xy/^
{3x + 2)
"^3/2
■~—{3x + 2) I = 0.353 = 35.3%
2 Jo.50
(b) P,
'0.50. 0.75 ~ 2 (3j: + 2) I
*_ (1 - fc)^/^
(3Zj + 2) + 1 = 0.5
(1 - fc)3/2(3fc + 2) = 1
fe = 0.586 = 58.6%
79. p = 1.20 + 0.04r
„ 15.000 f'"' ,
c = —ri-\ Pds
(a) 2000 corresponds to f = 10.
C = -^f"[l.20 + 0.04r]dr
^ Jio
1.20r + 0.02f'
^ i5,ooor
~ M [
" 24,300
10 M
(b) 2005 corresponds to f = 15.
C =
15,000
M
1.20r + 0.02r^
"^ _ 27,300
15 ~ M
81. Trapezoidal Rule (« = 4): J_ -p^^ »= |[y^ + l + (',25)3 + TtW ^ 1 + (^75)3 + TT2^
Simpson's Rule (n = 4): I , dx " ~r7:\~r
+ -4^ + - '^
+ 1' 1 + (1.25)3 1 + (1 5)3 1 + (J 75)3 1 + 2
= 0.257
0.254
Graphing utility: 0.254
83. Trapezoidal Rule (,
« = 4):
Jo
^ COS xdx ~ 0.637
Simpson's Rule (n = 4): 0.685
Graphing Utility: 0.704
SS. (a.) R < 1 < T < L
(b) S(4) = ^^[/(O) + 4/(1) + 2/(2) + 4/(3) +/(4)]
4 + 4(2) + 2(1) + 4i^] + i
5.417
214 Chapter 4 Integration
Problem Solving for Chapter 4
1. (a) L(l)
=IV-
1
(b) L '{x) = - by the Second Fundamental Theorem of Calculus.
X
L\\) = 1 .
(c) L{x) = 1 = I -rfrfor;c « 2.718
-^n^
dt = 0.999896 (Note: The exact value of x is e, the base of the natural logarithm function.)
f
P'l f 1
(d) We first show that \ -dt = \ -dt.
To see this, let m = — and du = — dt.
Xy X^
P'l f 1 f 1 f 1
Then -dt= — (x, du) = - du = \ -dt.
h t JiA.^i JiA," JlA.f
/■jTAj Pm / f\
Now, i-(x,x-,) =1 -dt = \ —du\ using « = —
Ji t JiA,« V Xi/
f 1 Pn
Ji/x,« Ji "
P'l P^l
= -du+ -du
Ji " Ji «
= LUi) + LUj).
= IH't
-2--
3. 5(;c) = I sini ^ Uf
(a)
(b)
V2 V5 2 V? vS-y7 2Vl3
The zeros of y = sin ^r— correspond to the relative
extrema of S{x).
(c) 5'W = sin-^ = 0
= mr =^ X' = 2n => X = V2»2. n integer.
Relative maximum at jc = ~Jl = 1.4142 and x = V6 «= 2.4495
Relative minimum at x = 2 and x = ,/8 = 2.8284
.2
(d)5"(x) = cos(^)(7rx) = 0
= — + AjTT => x^ = 1 + 2n => X = Vl + 2n, « integer
Points of inflection at x = 1, V3, Js, and ^7.
Problem Solving for Chapter 4 215
5. (a)
(8,3)
(b)
x
0
1
2
3
4
5
6
7
8
1
7
7
1
Hx)
0
"2
-2
~2
-4
~2
-2
4
3
(c) f(x) =
-X, 0 < .r < 2
X - 4, 2<x<6
ij: - 1, 6 < ;c < 8
0 < -r < 2
F(x]
., [(-^72), ......
= /(f) dt = I (.x:V2) - 4x + 4, 2<A<6
■''' L(l/4):r2_ ^ _ 5 ^ < x < "
F'(x) = f(x). F is decreasing on (0, 4) and increasing on
(4, 8). Therefore, the minimum is — 4 at .r = 4, and the
maximum is 3 at .r = 8.
(d) F"{x)=f\x) =
f-l, 0 < Jt < 2
1, 2 < .r < 6
J.,
2
6 < j: <
At = 2 is a point of inflection, whereas x = 6 is not.
(/is not continuous at j: = 6.)
cos.ri±c « cosi ^1 + cosi — ?= j = 2 cosi — p 1 = 1.6758
7. (a) cos.ri±c « cosi — -i= \ + cos'
I cos a: dx: = sin X = 2 sin(l) = 1.6829
Error. 11.6829 - 1.6758 = 0.0071
(b)
r.i
1
; dx ~
1
1
1 +A-2 1 + (1/3) 1 + (1/3) 2
(Note: exact answer is 7r/2 == 1 .5708)
(c) Let p(x) = ax^ + bx- + ex + d.
b.^ cx-
+ —- + —- + <ir
3 2
1
lb
3
= =r + Id
'^--ir'\ji
.,|..).g..).|..
9. Consider Hv) = [fix)]- => F\x) = 2f{x)f\x).'\\ms,
\f{x)f\x)dx= \\
Ja Jo ^
F'(x)dx
Fix)
= ^[F(b) - F(a)]
= pibT- - f(ar-]
11. Consider x^ dx =
Jo
The corresponding Riemann Sum using right endpoints is
Si
"' = e
' - 1^ ^
= \[l^ + 25 + ■ ■ ■ + n5]
Thus, lim S{n) = lim
1^ + 25 +
~ 6'
216 Chapter 4 Integration
13. By Theorem 4.8, 0 < /W < M => \ f{x) dx < I M dx = M(b - a).
Ja Ja
rb cb
Similarly, m < f(x) => m{b — a) = mdx < f(x) dx.
Ja Ja
Thus, m{b - a) < \ f{x) dx < M{b - a). On the interval [0, 1], 1 < JT^H? < V2 and fo - a = 1.
Ja
Thus, 1 < J\ + x^dx < 72. (Note: I yr+~?c&= 1.0894J
15. Since-|/W| < f{x) < \f{x)\,
J'b rb rb rb rb
\f{x)\dx< \f{x)dx< \f{x)\dx ^ \ f{x)dx <\ \f{x)\dx.
a Ja Jo Ja Jo
1 r
100,000
. l-nit - 60)1 _, 100,000
1 + sm r— 7 dt =
36.5
'\
365
365 l-nit -
t - ^r-COS -
2lT
t - 60)T' ^
365 Jo
100,000 lbs.
CHAPTER 5
Logarithmic, Exponential,
and Other Transcendental Functions
Section 5.1 The Natural Logarithmic Function: Differentiation .... 218
Section 5.2 The Natural Logarithmic Function: Integration 223
Section 5.3 Inverse Functions 227
Section 5.4 Exponential Functions: Differentiation and Integration . . 233
Section 5.5 Bases Other than e and Applications 240
Section 5.6 Differential Equations: Growth and Decay 246
Sections.? Differential Equations: Separation of Variables 251
Section 5.8 Inverse Trigonometric Functions: Differentiation 259
Section 5.9 Inverse Trigonometric Functions: Integration 263
Section 5.10 Hyperbolic Functions 267
Review Exercises 272
Problem Solving 278
CHAPTER 5
Logarithmic, Exponential, and Other Transcendental Functions
Section 5.1 The Natural Logarithmic Function: Differentiation
Solutions to Odd-Numbered Exercises
1. Simpson's Rule: /j = 10
X
0.5
1.5
2
2.5
3
3.5
4
Jl '
-0.6932
0.4055
0.6932
0.9163
1.0987
1.2529
1.3865
Note:
Jl t Jos t
dt
3. (a) In 45 » 3.8067
ri
(b) I -dt == 3.8067
\y
5. (a) In 0.8 = -0.2231
ro.8
(b) I -dt == -0.2231
ro.8
-dt == -0.:
Jl t
7. f{x) = \nx + 2
Vertical shift 2 units upward
Matches (b)
11. f{x) = 3 in ;c
Domain: x > 0
13. fix) = In 2x
Domain: jc > 0
9. fix) = In U - 1)
Horizontal shift 1 unit to the right
Matches (a)
15. fix) = In(x - 1)
Domain: x > \
17. (a) In 6 = In 2 + In 3 = 1.7917
(b) Inf = In2 - ln3 = -0.4055
(c) In 81 = In 3'' = 4 in 3 « 4.3944
(d) In 73 = in 3'''2 = i |n 3 « 0.5493
19. Inf = In 2 - In 3
21. In — = In X + In y - In z
23. In 7a2+ 1 = ln(a2 + i)i/3 = - in(a2 + 1)
218
Sections.] The Natural Logarithmic Function: Differentiation 219
25. Inf^^-T-^V = 3[ln(;c2 - 1) - In.r^]
x'
= 3[lnU- + 1) + InU - 1) - 3 In.x]
27. In z(z - 1)2 = In z + ln(z - 1)^
= In z + 2 ln(z - 1)
29. ln(;c - 2) - InU + 2) = In
x-2
x + 2
— 1 r» , / ,x , , / 1 .\T 1 , -tU + 3)- , , /x(x + 3)2
31. -[21ii(a: + 3) + Inx - InU^ - 1)] = - In ^^, _ ^ = In ^ '^^ _ ^
33 . 2 In 3 - - \n(x- + 1) = In 9 - lnVx= + 1 = In , ^
2 Vj;' + 1
35. 3
37. lim ln(x - 3) = -oo
j-»3*
39. lim ln[x2(3 - x)] = In 4 - 1.3863
X — »2
41. y = In j:^ = 3 In .r
, 3
At(l,0),v' = 3.
43. >' = In JT' = 2 In ;c
At(l,0),v'= 2.
45. g(x) = Inx- = 21nA;
47. V = (ln;c)''
^ = 4(lnx)3(i)=^(ill^
ax \x/ X
49. V = In xjx^ - 1 = In .X + - In(x2 - 1)
_ 2.r2 - 1
dc X 2\x2 — 1/ xU' — 1)
^ = i + V ^
51. /W = In^— - = In.T - In(.r- + 1)
„, . 1 2x l-x-
/ w = r
x x^ + 1 x(x^ + 1)
53. g{t) =
Int
g'it)
t-(l/t) - 2t\nt 1 - 2Inr
55. .v = In(Inx-)
^ = -!-— (In.r2) = (^^'/-^^ = - = 1
djc In x^ a[r ' In x- x In x- x In x
57. y = \n
yj^ = f[ln(x+l)-ln(x-l)]
dy ^ J_r_l 1_1 ^ 1
dx 2Lx + 1 X - ij ~ 1 -x2
59. f(x) = In
J 4 + x^ 1
-ln(4 + x^) - Inx
fix)
4 + x^ X x(x~ + 4)
220 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
61.
y =
■v^c^+T
+ ln(;c + Jx- + 1 )
dl^ -x[x/Jx- + l) + y/jC^ + 1
dx~ £-
1
1
1 +
x + -Jx- + 1 / V V;c2 + 1
1
J^fT\+ .
x^J^^^Tx \x + JWT\j \ J^T\ I x^JlFTl 7]?TT ^-J^FTx
+ ■
1 + x- v^2~rT
63. y = In I sin .r|
dV _ cos.T
dx sinjc
cotx
65. y = In
COSJC
COS j: - 1
= Inlcosxl - Inlcosj: — ll
-sin a:
dy _ — sinx _
dx cos X cos X — I
— tan.r +
sin X
cosj: - 1
67. >> = In
■ 1 + sin x
2 + sinx
= ln| — 1 + sin xl — ln|2 + sinx|
dy _ cos X cos x
dx — 1 + sin .t 2 + sin X
_ 3 cos X
(sinx - l)(sinx + 2)
69. fix) = sin 2x In x^ = 2 sin 2x In x
fix) = (2 sin 2x)(-J + 4 cos 2x in x
= - (sin 2x + 2x cos 2x In x)
X
2
= ^sin 2x + X cos 2x In x^)
71. (a) >- = 3x2- In X, (13)
dy I
— = 6x
dx x
. When X = 1, v" = 5.
dx
Tangent line: y - 3 = 5(x - 1)
y = 5x - 2
0 = 5x - 3^ - 2
(b)
73. x2 - 31n>' -l-y^ = 10
y dx dx
-le-'
dy _ 2x 2x3;
dx " i-i/y) -ly~ l-ly^
75. y = 2(lnx) + 3
77. y = j-Inx
Domain: x > 0
1 (x + l)(x - 1)
>> =x
0 when x = 1 .
1
/'=l+^>0
Relative minimum: ( 1 , 2)
^
Section 5.] The Natural Logarithmic Function: Differentiation 221
79. y = xlnx
Domain: ;c > 0
y' = xl-] + \nx = 1 + In .r = 0 when x = e~K
y"=- > 0
Relative minimum: (e ', — e ')
(f-'.-f-')
81. V =
Injc
Domain: 0 < jc < l,.x > I
, (in;c)(l) - (x){l/x) _ In-r- 1
y =
= 0 when x = e.
(ln;c)2 (Inx)'
(ln;c)^(lA) - (Inx - l)(2/;c) In a: 2 - \nx
{\nxr
Relative minimum: (e, e)
Point of inflection: {e~, e^/2)
x(\nx)^
0 whenx = e^.
83. fix) = ln;c, /(I) = 0
/'W=^, /'(1)=1
/"W = -A, /"(i) = -i
PiW =/(l) +/'(1)U - 1) = X - 1, P,(l) = 0
/'2W =/(l) +f'Wix - 1) + |/"(1)U - IP
= U-l)-^(x- 1)% P,(1) = 0
P,'(.r) = 1, P/(l)= 1
P/W = 1 - (x - 1) = 2 - .r, P.'(l} = 1
M;c)= -1, P/i;i)= -1
85. Find.x such that In.r = — a:.
fix) = ilnx)+x = 0
1
fix) = - + 1
" fix J -''■
1 - ln,T„
1 +.v„
M
1
2
3
^.
0.5
0.5644
0.5671
/UJ
-0.1931
-0.0076
-0.0001
The values of/, Pj, Pj, and their first derivatives agree at
X = 1. The values of the second derivatives of/ and P,
agree at j: = 1 .
87. y = x^x^ - 1
Inr = In.v + |ln(.t= - 1)
y\dxj X ;c2 - 1
dt \xix^ - 1)J ^/p^H"
Approximate root: x = 0.567
222 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
89.
y =
1
]ny = 2 ln;c + -ln(3;c - 2) - 2 In^ - 1)
lf^U2^ 3
y\dx) X 2(3a- — 2) x — I
dy
dx
y
3x^ - \5x +
2xOx - 2){x - 1).
3x!^ - 15;>:^ + 8jc
2{x - iyj3x - 2
91.
1)3/2
V7+T
3 1
Iny = \nx + - InU - 1) - - ln(;t + 1)
3/ 1
U 1
y\dxj X 2\x — 1/ 2\x +
dy
dx
2Lx
X + 1
y^4x^ + \x- 2
21 xix''- 1)
{2x^ + 2x- \)Jx- 1
{x + 1)3/2
93. Answers will vary. See Theorems 5.1 and 5.2.
95. In e' = ;c because f{x) = In x and g(x) = e^
are inverse functions.
97. (a) /(l) 9^/(3)
(b) fix) = 1 - - = 0 forA: = 2.
X
99.
/3 = 10 log,,
10\ 10-16
l«lnf ^
In 10 VIO
10
In 10
{In/ + 161n 10] = 160 + lOlogig/
j8(10"'°) = 7^[ln 10-'° + 16 In 10] = 7^[- 10 In 10 + 16 In 10] = 7^[6 In 10] = 60 decibels
1 In 10 m 10 In 10
101. (a) You get an error message because In h does not exist
for /z = 0.
(b) Reversing the data, you obtain
h = 0.8627 - 6.4474 In p. ■ '
[Note: Fit a line to the data (x, y) = (Inp, h).]
(C) 25
(d) If p = 0.75, h == 2.72 km.
(e) If h = 13 km, p = 0.15 atmosphere.
(f) /! = 0.8627 - 6.4474 In/?
1 = -6.4474--^ (implicit differentiation)
p dh
dp ^ p
dh -6.4474
For A = 5, p = 0.5264 and dp/dh = -0.0816 atmos/km.
For h = 2Q),p = 0.0514 and dp/dh = -0.0080
atmos/km.
As the altitude increases, the rate of change of pressure
decreases.
103. (a) fix) = In x, gix) = J'x
{h) fix)^\nx,gix)=ifx
^■'"4-'«^i^
For X > 'X, g 'ix) > fix), g is increasing at a faster rate
than/for "large" values of z.
fix)
For X > 256, g Xx) > fix), g is increasing at a faster rate
than/for ' ;arge" values of .v./(.v) = In x increases very
slowly for "large" values of x
Section 5.2 The Natural Logarithmic Function: Integration 223
105. False
In j: + In 25 = ln(25x) # ln(;t + 25)
Section 5.2 The Natural Logarithmic Function: Integration
BH
-dx = 5\n \x\ + C
3. « = .r + 1, rf« = dx
1
J.r + 1
dx = InLc + 1 + C
5. u = 3 — 2x, du = —2dx
^ dJc=-l-\—^{-2)dx
Js-Zx'^ 2j3 -2x
1
In 3 - 2x\ + C
7. u = x^ + I, du = 2x dx
-ln{x2+ 1) + C
InV.r^ + 1 + C
/^-/h!
ott
- 4ln|;c| + C
11. M = .r^ + 3x- + 9x, du = 3(x~ + Zx + 3) dx
r .t- + 2x + 3 ^ 1 f3{x^ + 2j: + 3)
J x3 + 3jc- + 9.t: " 3J .r^ + 3a2 + 9.r
■ ln|x3 + 3x^ + 9x\+ C
..j^f^.-j{.-^^jh].
4x + 61nU + 1| + C
"■/^^:^*^/(''--?tiK
y-2;c + |ln(;c2 + 2) + C
21. H = a; + I, du = dx
= Ux+ l)-'/2(ic
= 2{x + 1)1/2 + C
lvlTT* = l<
= 2V.r + 1 + C
-/^4^-/('-7^)-
= Y + 51n|;c - 3| + C
19. M = in X, du = - dx
X
\
{\nxf , _ 1„_.,3
dx = -(ln.r)' + C
X 3
f 2r flv - 2 + 2
=/n--/u^-
= 21nU - 1|
(.V - 1)
+ C
224 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
1
25. M = 1 + Jlx, du = -^= dx =^ {u - \)du = dx
1
}\ + Jlx J
Jlx
■du
du
1 + V2jc / "
= « — ln|M| + C,
= (l + 72I) - Injl + JYx\ + Ci
= Jlx- ln(l + v^) + C
where C = Cj + 1 .
27. « = V^ - 3, rfw = -pdx => 2(m + 3) du = dx
ijx
\-^^dx = 2j^.« = l\^^^^^du = 2j(« . 6 + f).«
= 2
+ 6m + 91ni«|
Ci = M^ + 12« + 181n|M| + C,
= [Jx - 3f + 12(v^ - 3) + 18 \^Jx - 3| + C,
= ;»:+ 6V5 + IBlnlv^ - 3! + C where C = C, - 27.
f
29. P^rfe=ln|sine| +C
sin 6
(m = sin Q, du = cos 6 dS)
31. CSC 2j: ^ = - (
I CSC 2j: ^ = - I (esc 2.i;)(2) dx
■ Inlcsc 2j: + cot 2x\ + C
33. f-^^^
J 1 + sin r
37.. = |:
df = Inll + sinil + C
2- X
■dx
= -3
-dx
(1.0)^
i
:^
?V:^
x-2
= -31n|;f - 2| + C
(1,0): 0 = -31n|l - 2| + C ==* C = 0
y = -3 1n|;<;- 2\
,_ , secxtanx . , 1 , ,"
35. i r dx = Inlsec x - 1 1 + C
f
39. i = tan(2e) rf0
1
(0.2), 4
^1'
tan(2e)(2rf0)
A I I A ,
-ln|cos2e| + C
(0, 2): 2 = -- ln|cos (0)| + C =J. C = 2
i = --ln|cos2e| + 2
41,
dx X + 2
(a)
(0,1)
1=^*
(b) y
X + 2
>.(0) = 1 ^ 1 = In 2 + C
rfx = ln|x + 2| + C
C =
Hence, y = ln|x + 2| + I - In 2 = In
- in 2
X + 2
Section 5.2 The Natural Logarithmic Function: Integration 225
43.
fl^'^^tf'^l^^^'ll
= -In 13 « 4.275
45. M = 1 + \n X, du = — dx
X
i
(1 + \nxY
dx =
^1 +ln.r)3
-ff^-/>
X + 1
dx
-x^- X- \n\x+ 1|
-In 3
49.
f-
cos 6
sin 0
rfe
ln|e - sin e\
= In
2 - sin 2
1 — sin 1
« 1.929
51. -Inlcosjcl + C = in
cos.r
+ C = Inlsecr + C
53. Inlsec^c + tan.r + C = In
= In
(sec X + tan .r)(sec j: — tan jc)
(sec X — tan x)
+ C = In
sec'.t - tan-^.x:
secx - tanj:
+ C
1
sec X - tan x
+ C = — Inlsecr — tan j: + C
. I ^a!j; = 2(1 + v^) - 21n(l + v^x) + C,
= 2[Vx - ln(l + Jx)] + C where C = C, + 2.
57. cos(l - x)dx = -sin(l - .t) + C
p/2 r
59. I (cscr — sinjf)fl[x = — ln|cscjc + cot.i:| + cosj:
Jtt/A L
V2
72
ir/4
= ln(V2 + 1) - -^ « 0.174
Note: In Exercises 61 and 63, you can use the Second Fundamental Theorem of Calculus or integrate the function.
=B
61. F(x) = \ -dt
F'ix) = i
62.fix) = [\dt = ['\dt-[\dt
F'ix) = f - i = 0
3a' .V
65.
A = 1.25
Matches (d)
67. A = I '^^^-^dx = I (x + -]dx
'\^
+ 4 In .r
' = (8 + 41n4)-^
15
— + 8 In 2 ~ 13.045 square units
*^
226 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
69. r2sec^d^ = -r
Jo 6 ttJo
ttxXtt ,
12
In
;— + tan —
6 6 I Jo
11
77
12
In
sec— + tan— ln|l + 0|
3 3 77
ln(2 + 73) - 5.03041
71. Power Rule
73. Substitution: (m = j:^ + 4)
and Log Rule
75. Divide the polynomials:
= j: — 1 +
x+ 1
X + 1
J p g ("4
77. Average value = r- ~dx = 4 x'''^dx
4 - 2J2 X- J2
-4i
-M
79. Average value
^tI'
Inx , 1
dx =
X e — 1
\_nxir
1 /I
e - 1V2
1
2€ - 2
= 0.291
81. Pit) = r fo25/' " (3000)(4)J^ "p^-^g/? = 12,000 In |1 + 0.25r| + C
P(0) = 12,000 ln|l + 0.25(0)1 + C = 1000
C = 1000
P(t) = 12,000 ln|l + 0.25f| + 1000 = 1000[121n|l + 0.25f| + 1]
P{3) = 1000[12(ln 1.75) + 1] « 7715
83.
50
^r3Sfi-=HH~
+ 3x\
_ 40
« $168.27
85. (a) 2x-- y^ = S
• ■ ^2 = 2x2 - 8
y, = V2x2- 8
:y2 = - V2x2- 8
\
^
/
'\
Let/: = 4andgraphr.^ ^ I - if^)
(c) In part (a), 2^^ - / = 8
4jc - 2yy ' = 0
, 2x
In part (b), y^ = - = 4j: '
x
2yy' =
,'_-^ _ ~2y _ -2y _ -y
yx^ y'^ x'^ 4x 2x'
Using a graphing utility the graphs intersect at (2.214, 1.344). The slopes are 3.295 and -0.304 = (- l)/3.295, respectively.
Section 5.3 Inverse Functions 227
87. False
I
(lnx) = ln(;c'^2)^(,n^)i/2
Section 5.3 Inverse Functions
89. True
"l
il
dx = InU + C,
= ln|x| + ln|C| = ln|C;c|, C ^0
1. (a) f(x) = 5;c + 1
/(gW) =/(^) = 5(^1 + l=x
giifU)) = g(5x + 1) = ^^""^P ' = X
3. (a) fix) = x'
g(x) = ^x
f{g{x))=f[ifx) = {irxy = x
g(/W) = g(:^)= W = x
(b)
I _2--
-3
5. (a) fix) = vT^^
gU) = x= + 4, .V > 0
/(gW)=/U- + 4)
= VU^ + 4)-4=v^ = ;c
= (y;c -4)2 + 4 = x-4 + 4
(b)
12--
10-
4
2
-\ 1 1 l-»-J
2 4 6 8 10 12
7. (a) fix) =
1
g{x) = -
f(g{x)) = Y/^ = ^
g(/W) = tV = ^
l/x
(b)
H 1 1—*--'
9. Matches (c)
11. Matches (a)
228 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
13. fix) = ja: + 6
One-to-one; has an inverse
15. f{e) = sin d
Not one-to-one; does not have an
17. h(s) =
1
s-2
One-to-one; has an inverse
19. /W = \nx
One-to-one; has an inverse
21. g(x) = U + 5)3
One-to-one; has an inverse
-2
23. f(x) -- {x + af + b
f'(x) = 3(;c + a)2 > 0 for all x.
/is increasing on (— oo, oo). Therefore, /is strictly
monotonic and has an inverse.
25. /W = ^ - 2x^
f'ix)=x?-4x = 0 when ;c = 0, 2, -2.
/is not strictly monotonic on (— oo, oo). Therefore, /does
not have an inverse.
27. /(x) = 2 - X - ;c3 . -.
fXx) = - 1 - 3;c2 < 0 for all x.
/is decreasing on (— oo, oo). Therefore, / is strictly monotonic and has an inverse.
29. fix) = lx--b=y
x = '-^
2
X + 3
y =
/-'W =
2
X + 3
31. /W = x5 = y
x= 4/y
y= Vx
/-'(x) = 5/^ = x'/5
33. /(;c) = v^ = y
X = y^
y = x'-
f-\x) = jc2, jc > 0
Section 5.3
Inverse Functions
229
35. fix) = JA- x'^ = y, 0 <x <2
x= J A -y^
= JV^J^
f-\x) = V4 - x\ 0 < ;c < 2
37. fix) = Vx- 1 = y
jc = y3 + 1
y = x^ + \
/-'W =x3+ 1
r
^'r
4-
^
The graphs of /and/"' are
reflections of each other
across the line y = x.
39. fix) = x^l^ = .V, ;( > 0
x = y3/2
f'\x) =;c3/2, ;c > 0
The graphs of/and/"' are
reflections of each other
across the line y = x.
41. /(x)=^#= = y
77;c
-^-yr^
/-
2
f::^
-^z
Jly
1 < ;<: < 1
The graphs of/ and/ ' are
reflections of each other
across the line v = x.
43.
X
1
2
3
4
f-\x)
0
1
2
4
2 3 4
45. (a) Let x be the number of pounds of the commodity
costing 1.25 per pound. Since there are 50 pounds
total, the amount of the second commodity is 50 — x
The total cost is
V = 1.25.r + 1.60(50 - x)
= -0.35a: + 80 0 < .r < 50.
(b) We find the inverse of the original function:
V = -0.35;c + 80
0.35a: = 80 - >>
X = f (80 - y)
Inverse: y = -^(80 - x) = f (80 - .x).
.r represents cost and y represents poimds.
(c) Domain of inverse is 62.5 < .v < 80.
(d) If .t = 73 in the inverse function.
y = -^(80 - 73) = -x = 20 pounds.
230 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
47. fix) = (x - 4)2 on [4, oo)
f'(x) = 2U - 4) > 0 on (4, oo)
/is increasing on [4, oo). Therefore, /is strictly
monotonic and has an inverse.
49. f(x) = -^ on (0, oo)
fix) = — r < Oon(0, oo)
x'
/is decreasing on (0, oo). Therefore, /is strictly
monotonic and has an inverse.
51. f(x) = cos X on [0, tt]
f'(x) = — sin x < 0 on (0, it)
/is decreasing on [0, it]. Therefore, /is strictly monotonic and has an inverse.
53.
fix)
x^--4
X^y — 4y = X
x^y — X — 4>' = 0
a = y, b = — 1, c = —4y
yon (-2, 2)
^ 1 + Vl - 4(y)(-4v) ^ 1 ± Vl + 16>-2
""' _ 2y 2> .
f(l - WTj6?)/2x, ifx ¥= 0
0, if X = 0
y=r'ix)
Domain: all x
Range: -2 < y < 2
^
^
The graphs of /and/"' are
reflections of each other
across the line >> = x.
55. (a), (b)
/
/-'
-fl
-^
(c) Yes, /is one-to-one and has an inverse. The inverse
relation is an inverse function.
57. (a), (b)
(c) g is not one-to-one and does not have an inverse.
The inverse relation is not an inverse function.
59. /U) = Jx - 2, Domain: x>2
fix)
> OfoTx > 2.
2yr^
/is one-to-one; has an inverse
Jx - 2 = y
x-2=f
x = f + 2
y = x^ + 2
f'^ix) = x^ + 2,x > 0
61. fix) = |x- 2|,x < 2
= -(x-2)
= 2 -X
/is one-to-one; has an inverse
2 — X = >>
2 — y = x
f~\x) = 2 - X, X > 0
63. fix) = ix - 3)2 is one-to-one for x > 3.
(x - 3)2 = >-
X - 3 = v^
x= Vy + 3
y=Vx + 3
/-'(x) = ^ -I- 3, X > 0
(Answer is not unique)
65. /(x) = |x -I- 3| is one-to-one forx > -3.
X -I- 3 =y
X = y — 3
y = X — 2
/-H3c) = X - 3, X > 0
(Answer is not unique)
Section 5.3 Inverse Functions 231
67. Yes, the volume is an increasing function, and hence
one-to-one. The inverse function gives the time /
corresponding to the volume V.
69. No, C{t) is not one-to-one because long distance costs are
step functions. A call lasting 2.1 minutes costs the same as
one lasting 2.2 minutes.
71.
fix) = :c^ + 2x - 1, /(I) = 2 = a
fix) = 3.r2 + 2
if-r{2)-j;rj^
1
1
1
/'(/-'(2)) /'(I) 3(iP + 2 5
73. /W = sin;t,/(|)=| = a
(/-
fix) = cos .r
1
1
2/ /'(/-'(1/2)) /V/6) cos(7r/6)
1 _ 2V3
73/2 3
75. /W = ^ - -■ /(2) = 6 = a
fix) = 3.r= + ^
(J ) yo) f'( f~^((i\\ f'(i\
1
1
/'(/-H6)) /'(2) 3(2)2 + (4/22) 13
77. (a) Domain/ = Domain/ ' = (—00,00)
(b) Range/= Range/"' = (—00,00)
(c)
3-1-
//
(d)
/(.) = ^, (|,^
/I
'X 1^1
/u
/-H;c)
(/-')'(.r) =
(/-
</.t-
79. (a) Domain/ = [4, 00), Domain /^^ = [0. oc)
(b) Range/ = [0, 00), Range/"' = [4, 00)
(c)
4 6 8 10 12
(d)
fix) = TT^, (5, 1)
Ax) =
2jx - 4
/'(5) = ^
/"'(x) = X- + 4, (1,5)
(/"■)'(.r) = 2x
(/-')'(1) = 2
232 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
81. .r = y3 - V + 2
^ ^d\ , , dy
dv
' Mi-4A),^ '
1
dx liy- - 14v' ' "ate 3-14 11
Alternate solution; let /(a) = x^ - Ix^ + 2.
Then/'(jc) = Sjc^ - 14Aand/'(l) = -11.
In Exercises 83 and 85, use the following.
/(x) = |a: - 3 andg(x) = x^
r\x) = %{x + 3) and g-\x) = i^
83. (/-' "g-')(l) =/-'(g-Hl)) =/-'(!) = 32
85. (/-' »r')(6) =/-'(/-'(6)) =/-'(72) = 600
In Exercises 87 and 89, use the following.
f(x) = X + 4 and gix) = 2* — 5
r'W=x-4andg-'W = ^
87. (g-' ./-i)W = r'(/-'W)
= g-'U - 4)
^(x- 4) + 5
2
x+ \
89. (/°g)(x)=/(gW)
= /(2x-5)
. = (Zx - 5) + 4
= 2x- 1
Hence, (/"g)-'U) =
X + 1
(Note: (/-g)-' =g-'"/-')
91. Answers will vary. See page 335 and Example 3.
93. y = x-^ on {— CO, ca) does not have an inverse.
95. /is not one-to-one because many different x- values yield
the same >'-value.
Example: /(O) = /(tt) = 0
(2rt — IItt
Not contmuous at , where n is an mteger
97. Let if'gjix) = y then x = (/■= g)^'(y). Also,
{f'g)ix)=y
figix)) = y
g(x)=r'(y)
X = g-'{f-'(y))
= ig-' 'f-')(y)
Since/and g are one-to-one functions,
(/°5)-' = g^'°r'-
99. Suppose g(x) and h{x) are both inverses of/(x). Then the graph of/(x) contains the point {a, b) if and only if the
graphs of gix) and h{x) contain the point (b, a). Since the graphs of g{x) and h(x) are the same, g{x) = h{x).
Therefore, the inverse of/(x) is unique.
Section 5.4 Exponential Functions: Differentiation and Integration 233
101. False
Let/W = x\
105. Not true
X, 0 < ;c < 1
1 - jc, 1 < X < 2 ■
Let/W
/is one-to-one, but not strictly monotonic.
103. True
107.
fix)
(/-')'(o)
dt
71 + t
1
ym^
/'(2) i/yi?
;, /(2) = 0
= yi7
Section 5.4 Exponential Functions: Differentiation and Integration
1. e° = \
Inl = 0
3. In 2 = 0.6931
g0.6931. =2
5. e'"-' = 4
X = 4
7. e' = 12
JC = In 12 = 2.485
9. 9 - 2?* = 7
2^^= 7
e' = 1
x = 0
11. 50e-^ = 30
-X = In
-4
0.511
15. InU - 3) = 2
X — i = e-
.t = 3 + e2 == 10.389
13. In .r = 2
.r = e= - 7.3891
17. ln7x + 2 = 1
Jx + 2 = e' = e
X + 2 = e-
jt = e^ - 2 = 5.389
19. V = e
21. >• = e-"^
Symmetric with respect to the .%-axis
Horizontal asymptote: v = 0
234 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
23. (a)
^
V
Horizontal shift 2 units to the
right
(b)
/f
k
A reflection in the jc-axis and a
vertical shrink
(c)
^
Vertical shift 3 units upward
and a reflection in the y-axis
25. y = Ce'"
Horizontal asymptote: v = 0
Matches (c)
27. y = C(l - €-<")
Vertical shift C units
Reflection in both the x- and y-axes
Matches (a)
29. f(x) = e^"
g(x) = InV^ = -\nx
31. f(x) = e--l
g(x) = ln{x + 1)
33.
;
g
^
As j: — > oo, the graph of /approaches the graph of g.
lim (1 +
X — »oo
^r
35. 1 +
1
1,000,000
1.000,000
« 2.718280469
e «= 2.718281828
e > 1 +
1
1,000,000
37. (a) y = e^^
At(0, l),y'= 3.
39. f{x) = e^
f'(x) = 2e^
45. g(i) = (e-' + e')^
g'W = 3(e-' + e')V-e-')
(b) y = e-
41. /(jc) = e-2'+^'
f-"
47. >' = In<r''
Ik-
y' = -Ze-^"
At(0, l),.v' =
• -J
.
43.
y =
dy _
dx
Ij-x
49.
y =
di_
ln(l +
2e^
e^)
dx I + e^
Section 5.4 Exponential Functions: Differentiation and Integration 235
51. y =
e' + e-
2{e^ + e-")'^
dy
dx
= -lie' + e-'Y^e' - e'")
-2(e^ - e'')
{e' + e-^)2
53. y = x^e^ - Ixe' + le' = e^ix^ - 2x + 2)
^ = e^(2x - 2) + e'ix^ - 2x + 2) = .tV
55. /(;c) = e-' In j:
Z'U) = e'4-] - e-'\BX = e~4- - In^J
57. y = e'isin x + cos x)
—- = e^icQis x — sin x) + (sin x + cos xSie' )
dx
= e^(2 cos j:) = le' cos at
59. xey - mx + 3.V = 0
xey^- + ey - 10 + 3^ = 0
dx dx
dy
dx
(xey + 3) = \0- ey
dy _ 10 - gy
dx " xey + 3
61. fix) = (3 + 2x)e-^'
fix) = (3 + 2x)(-3e-3') + le-^"
= (-7 - 6A;)e-3^
/W = (-7 - 6x){-3e-^') - 6e-3x
= 3(6;c + 5)e-^'
63. J! = e^{cosV2jc + sinv/2Jc)
>>' = e^(- V2sin ^x + V2cos V2;>:) + e-^fcosVlv + sin v^)
= ej{l + 72)cosV2x + (l - >/5)sin>/2»:]
>>"= ej_-{j2 + 2)sin v/2x + (^2 - 2)cos J2x] + e^[(\ + V2)cos72Jc + (l - v^sin Vlt]
= e{(- 1 - 2V2)sin ^x + (- 1 + 2V2)cos V2a]
-2>'' + 3v = -2e-'f(l + V2)cos Vli: + (l - 72)sin Vlr] + 3e{cosV2.i: + sin Tlr]
= e{(l - 2v/2)cos y2.v + (l + 272)sin Vlx] = -y"
Therefore, -2y' + 3y = -y" => y" - 2.v' + 3.v = 0.
65. f(x) =
e^ + e-
f'(x) = ^ .^ " = 0 when x = 0.
fix) = ^-^-^ > 0
Relative minimum; (0, 1)
67. gW = -^,-U-2)V2
8'{x) = ^(.v - 2)g-(-2)V2
V2Tr
g"(x) = -^(.t - l)(.r - 3)g-(-2'V2
v/2^
1
Relative maximum: ( 2, — ;= ) ~ (2, 0.399)
'Z7T
1
Points of inflection: I 1,
^r- 1 3
,.-1/: ==
(1,0.242), (3,0.242)
236 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
69. fix) = x-e-'^
fix) = -jc^e--- + 2x6''^ = xe'^il - x) = 0 when jc = 0, 2.
fix) = -e''i2x - x^) + e-^(2 - 2x)
= e-^ix^ - 4x + 2) = 0 when;c = 2 ± V2.
Relative minimum: (0, 0)
Relative maximum: i2,4e~^)
x = l± Jl
y = (2 ± v^2^-(2±>^
Points of inflection: (3.414, 0.384), (0.586, 0.191)
°(2±V2,(6±4V2)c-"*^')
71. g{i) = 1 + (2 + t)e-'
. g'ii) = (1 + t)e-'
g"(i) = te''
Relative maximum: (—1,1 + e)
Point of inflection: (0, 3)
(-1,3.718)
{-\.\*e)
^0.3)
73. A = (base)(height) = Ixe'"'
dA
dx
= -Ax^e''^ + le-^
= 2e-^(l - Ix^) = Owhenx =
72
A = J2e~''^
75. y =
1 + ae-''"'
-x/b
.A-i'
,a>0,b>0,L>0
at
,-xlb
y =
(1 + ae-^l»Y (1 + ae-^I^Y
y"=
(1 + ae-''"'Y
(1 + ae-'/>')\~e-''^\ + 2(y <?--/* V^g-V*
(1 + ae-'''>'f
^ Lae-^"'[ae-''/'> - 1]
(1 + ae-'/^f fc2
y"= Olfae-"'" = 1
L
—r- = In - => j: = fo In a
b \aj
yib In a) =
1 +ae-(i'in«)/* 1 +a(l/a) 2
Therefore, the y-coordinate of the inflection point is L/2.
Section 5.4 Exponential Functions: Dijferentiation and Integration 237
77, e"^ = X =^ fix) = X- e'"
fix) = 1 + e"^
X, = 1
X2 = x,- |r4 = 0.5379
79. (a)
' f'(x2)
0.5670
^4 = *3 - TTfH " 0-5671
/U3)
We approximate the root of /to be .1: = 0.567.
(b) When x increases without bound, l/x approaches zero,
and e'''-' approaches 1. Therefore, f(x) approaches
2/(1 + 1) = 1. Thus, /(.r) has a horizontal asymptote
at y = 1. As j: approaches zero from the right, l/x
approaches co, e^^' approaches 00 and/(.r) approaches
zero. As x approaches zero from the left, l/x approach-
es — 00, e'/^ approaches zero, and/(.r) approaches 2.
The limit does not exist since the left limit does not
equal the right limit. Therefore, x = 0 is a
nonremovable discontinuity.
81.
h
0
5
10
15
20
P
10,332
5.583
2,376
1,240
517
]nP
9.243
8.627
7.773
7.123
6.248
(a) 12
y = -0.1499/! + 9.3018 is the regression
line for data (h. In P).
(c) 12.000
gOh
(b) In P = ah + b
P = pnh + b —
P = Ce^, C = e"
For our data, a = -0.1499 and C = e"0'8 = 10.957.7
P = 10,957.7e-o'''99''
(d) ^ = (10,957.71)(-0.1499)e-''"'9'"
all
= - 1642.56e-o '«»'■
For/i = 5,^ = -776.3. For /! = 18,^ « -110.6.
ah all
83. fix) = e'^fio) = 1
fix) = \e^'\fXQ) = i
fix) = iw^/"(0) = \
P: jf^'^
P,(;f) = 1 + 2(^-0)=:^+ l.PilO) = 1
P,'W=|,P/(0)=|
P2W = 1 + |(-v - 0) + |(.x- - 0)^- = I + ^ + 1, P,(0) = 1
P/W = \, Pz'tO) = ^
The values of/, f ,, Pj and their first derivatives agree at x = 0. The values of the second derivatives of/ and P;
agree at .v = 0.
238 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
85. (a) y = e^
/
(c) y = e'
x^ x^
/
>
Qo) y = e
^2 = 1 + ^ + ( y
^ ^
/
87. Let u = 'bx, du = 5 dx.
/
e ' 5dx = e^ + C
91.
Jxe-^dx=-|J,
e-'^{-2x)dx = -|e"^ + C
89. Let u = -2x,du = -2dx.
ju^=4|U(-2)^=[4.-
1 e^ _ 1
93. I ^dx = 2 I e^(-^ W = 2e-^ + C
95. Let H = 1 + e"-', du = -e""^dj:
I ^ "^ dx = - \-—^-^dx = -ln(l + e-^) + C = ln( ^ . ) + C = a: - ln(e^ + 1) + C
J 1 + e"^ J I + e-' ^ ' \e^+ 1/
3 3
97. Let u = -, du = — :;dx.
X X-
/:f-4H-i)-
-igSAl
'I
{e^ - 1)
99. Letu= I - e^,du = -e'dx.
le'JT^^dx = - (1 - e'y'-(-e')dx
= -^(1 - e'f'^ + C
101. Let « = e^ - e"-', <^u = (e^ + e-'^)dx.
/
e* — e
-iic = In e"^ - e"^ + C
103. r J" dx= \5e-^dx- \e-^dx
= -T^"^ + e"-* + C
Lsinm^Qs 7^^ = _Lsin,rx(^
COS T7;c) lie
— rtSin 7j:r
+ c
Y
107. I e-^Xw.(e-^)dx= - [tan(e~^)](-e-"^)att
= ln|cos(e-^)| + C
Section 5.4 Exponential Functions: Differentiation and Integration 239
109. Let u = ax^, du = lax dx. (Assume a i= 0)
y = I xe"^' dx
= \xe^(
— e'^(2ax) dx = —e^ + C
2a j 2a
/i
111. fix) = \^(e' + e-^)dx = ]-(e- - e'^) + C,
/'(O) = C, = 0
"l
/w
/:
^{e' - e-')dx = -(e^ + e"^) + C^
/(O) = 1 + Cj = 1 => Co = 0
f(x} = -(C + e-')
113. (a)
(0, d;
(b) ^ = 2e-^/% (0, 1)
= \le-''^dx= -4 e
1
>-= \2e-''^dx= -A\e-''^\--dx
= -4e-^/2 + c
(0, 1): 1 = -4e° + C= -4 + C => C = 5
t
115. I £^dA:=e' =e5-l = 147.413
76
117. I xe-'^^'^dx =
I
.lp-^/4
= -le-^l^- + 2 « 1.554
119. (a) /(m - v) = e"-" = (£")(£-") =
(b) /(fcjc) = e^- = (e')^ = [/W?.
e" _ /(«)
(-6
121. 0.0665 e-o.oi39Cr-48)^^,
J48
Graphing Utility: 0.4772 = 47.727c
123.
e'dt >
Jo Jo
\dt
e^-\>x^!-e^>\+ X for.v > 0
125. /(.v) = e". Domain is (—00. 00) and range is (0. oc).
/is continuous, increasing, one-to-one. and concave
upwards on its entire domain.
lim e'' = Q and lim e'' = 00.
a — » - 00 j:— >oo
127. Yes. fix) = Ce^.Ca constant.
129. e ' >
Jo
e'^dx > 0.
240 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
131. fix)
\nx
(a) f'(x) = ^ — = 0 when x = e.
On (0, e),f'{x) > 0 => /is increasing.
On {e, oo),/'(jc) < 0 => /is decreasing.
(b) For e < /4 < S, we have:
In A InS
B\t\A > A\nB
InA^ > InS''
A« > B^.
(c) Since e < tt, from part (b) we have e'^ > tt"".
Section 5.5 Bases Other than e and Applications
■-(r
' -^(r
5. log^l = log2 2-3 = -
At fo = 6, y =
(if
_ 1
~ 4
At fo = 10, y =
10/7
- 0.3715
7. log, 1 = 0
9. (a) 23 = 8
11. (a) logioO.Ol = -2
\ '
log28 = 3
(b) 3-'=^
10-2 = 0.01
(b) logo^8= -3
0.5-3 = 8
l0g3T=-l
%-' = 8
13. y = 3^
X
-2
-1
0
1
2
y
1
9
1
3
1
3
9
\ 1 — >-»
15. v =
X
-2
-1
0
1
2
y
9
3
1
1
3
1
9
17. h{x) = 5^-2
JC
-1
0
1
2
3
y
125
X
25
i
5
1
5
/
4-
/
3-
/
2-
1
\
1-
w.
1 2
3
A
*
Section 5.5 Bases Other than e and Applications 241
19. (a) logio 1000 = X
W = 1000
.t = 3
(b) logioO.l =x
i(y = 0.1
JC= -1
23. (a) x^- x = log; 25
X- - X = logj 5- =
2
;c2 - X - 2 = 0
(x + 1)U - 2) = 0
X = -10R;c = 2
25. 32^ = 75
2jcln 3 = In 75
. = 1'"^5^,Q..
2 In 3
21. (a) log3;c= -1
3-' =x
I
-f = 3
(b) log2J: = -4
2-" = X
1
(b) 3a; + 5 = log2 64
3jc + 5 = log2 2* = 6
3.r = 1
27. 23--' = 625
(3 - .r)ln 2 = In 625
In 625
3 - x =
In 2
3 -
In 625
In 2
-6.288
29. |l+«ff^3
12rln|l +~) = ln3
f =
1
In 3
12 /, ^ 0.09
T+I2-
== 12.253
31. logjU - 1) = 5
.r - I = 2^ = 32
.r = 33
33. logj x~ = 4.5
;c2 = 3^-5
.t = ±v^^ ±11.845
35. gU) = 6(2'--)-25
Zero: x= -1.059
^
(-1.059.0)
/
\
X.^
37. /lU) = 32 logioU - 2) + 15
Zero: 5 = 2.340
242 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
39. f(x) = 4^
g{x) = log4;c
x
-2
-1
0
i
2
1
fix)
1
16
1
4
1
2
4
x
1
16
1
4
1
2
4
gW
-2
-1
0
i
2
1
41. /U) = 4^
/'(x) = (ln4)4-
43. >- = 5^-2
ax
45. g{t) = t^2'
g'it) = tHln2)2' + (2t)2'
= r2'(rln2 + 2)
= 2'ti2 + t\n2)
47. /i(e) = 2-9cos ttO
h'ie) = 2-s(-7rsin ttO) - (ln2)2-»cos trd
. = -2-^(In 2) cos -770+ TTSin Trd]
49.
^
^
logs -'^
1
xln3
51. fix) = log.
/'W
55. git)
8 'it)
- 10S2
X - 1
= 2 log2 x - logj ix - 1)
2
xln
1
2 (;t-l)ln2
X- 2
(In 2)a:(jc - 1)
lOlog^r 10 /In A
r ln4V r /
10
rr(lA) - Inrl
In 4
L f' J
10
r, ,„ .1
5
/2ln4'-
;Mn2
(1 -Inf)
53. y = logj V^^^^ = I log; ix^ - 1)
^ = 1 2;c ^ X
dx 2' [x- - l)ln5 ~ U^- l)ln5
57.
y = X
'./x
In y = - In j:
X
^ = ^(1 - Inx) = 2x(2/-'-2(l - ln;c)
59. y=ix-2Y*^
Iny = (x + l)ln(x - 2)
1
Kl)=<-«u-.
+ InU - 2)
^
^
X + I
-^ + Hx - 2)
x - 2
= U - 2)--
i^
+ InU - 2)
61.j3^^ = |^
+ C
Section 5.5 Bases Other than e and Applications 243
In 2 L 2j
2 In 2 In 4
5./.5-=^=-l|5
65. \x5'''~ dx= -]r\5 '\-2x)dx
1\5-
\ll In 5
-1
+ C
2 In 5
(5-^'-) + C
r 3^
J 1 +3-
67. I . : d[»;, M = 1 + 3-', c/m = 2(ln 3)3^ dx
1
2 In 3)3-
2 In 3
m
d[x
^ ln(l+3^) + C
2 In 3
69. ^ = 0.4^/3^ (0, \
dx \ 2
Jo.4^^^^. = 3[o.4'/'(|
dx
In 0.4
0.4^/3 _,_ (; = 3(in 2.5)(0.4K3 + ^
.V = 3 In 2.5(0.4)^/3 + _ _ 3 ]„ 2.5
(a)
3(1 - 0.4^3) 1
In 2.5 2
71. Answers will vary. Example: Growth and decay problems.
(b)
7
73. y
(8,3)»
• (2.1)
(1,0)
-»-\ 1 1 1-
X
1
2
8
y
0
1
3
(a) y is an exponential function of -v: False
(b) y is a logarithmic function of .r: True; v = log^.v
(c) X is an exponential function of v: True, 2* = x
(d) y is a linear function of x: False
2 4 6
75./(.^) = log,x ^ f'(x) =
six.
1
xln2
x' =^ g'(x) = x^(l + In.x)
[Note: Let y = g(x). Then: In y = In a-"^ = x In .r
— y = A' • — h In .V
y .V
y' = y(l + In.v)
y' = x'^(l +lnA) = g'(j:).]
^(a) = .ir => ^ '(.t) = 2x
Ma) = 2' =» A''(.v) = (In 2)2^
From greatest to smallest rate of growth:
gix), k{x), h{x)J(x)
77. C{t) = P(1.05)'
(a) C(10) = 24.95(1.05)'°
-$40.64
(b) ^ = P{\n 1.05)(1.05)'
at
j^
Whenf = 1: — = 0.051P
dt
When f = 8: — = 0.072P
dt
(c) ^ = (In 1.05)[P(1.05)']
= (In 1.05)C(r)
The constant of proportionality
is hi 1.05.
244 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
79. P = $1000, r = 35% = 0.035, t = 10
/ 0 035\"'"
A= lOOofl +^^)
A = 1000e<oo35)(>o) = 1419.07
n
1
2
4
12
365
Continuous
A
1410.60
1414.78
1416.91
1418.34
1419.04
1419.07
81. P = $1000, r = 5% = 0.05, ( = 30
0.05 yo"
A = 10001 1 + 1
A = lOOOeCo^'^o = 4481.69
n
1
2
4
12
365
Continuous
A
4321.94
4399.79
4440.21
4467.74
4481.23
4481.69
83. 100,000 = P^-oo^i => p = lOO.OOOe^
t
1
10
20
30
40
50
p
95,122.94
60,653.07
36,787.94
22,313.02
13,583.53
8208.50
85. 100,000 = P 1 +
OOSV^'
12
P = 100,000 1 +
0.05 V'^
12
t
1
10
20
30
40
50
p
95,132.82
60,716.10
36,864.45
22,382.66
13,589.88
8251.24
0 06V365)(8)
87. (a) A = 20,000( 1 + ^j » $32,320.21
(b) A = $30,000
(c) A = 8000 1
0.06V
,(365)(8)
+ 20,0001
365/
$12,928.09 + 25,424.48 = $38,352.57
0.06Y36s)(4)
365/
= $34,985.11
Take option (c).
(365)(8)
+ 1 +
0.06V
365/
(365){4)
+ 1
89. (a) lim 6.7e<-*8'>/' = e.le" = 6.7 million ft^
r-»oo
322.27
(b)
V-
-(48.1)A
V"(20) == 0.073 million ftVyr
V'(60) = 0.040 million ftVyr
91. y
300
3 + ne'0.062Sjc
(a) '<»
(b) If ;c = 2 (2000 egg masses), y = 16.67 == 16.'
(c) If y = 66.67%, then x = 38.8 or 38,800 egg masses.
(d) y = 300(3 + ne-oos^^o:)-!
,_ 318.756-°°^^^
^ (3 + 17e-oo625x)2
„_ 19.921875e-'"^^^(17g-°'^^^ - 3)
^ (3 + 17e-0"625^)3
17^-0.0625^ - 3 = 0 => ;c = 27.8 or 27,800 egg masses.
Section 5.5 Bases Other than e and Applications 245
93. (a) B = 4.7539(6.7744)'' = 4.7539e''»32^
(b) '20
(c) SV) = 9.0952e'»'3M
5 '(0.8) == 42.03 tons/inch
5 '(1.5) = 160.38 tons/inch
95. (a) f(t) dt = 5.67
Jo
I g{t) dt - 5.67
Jo
h(t) dt - 5.67
Jo
(b)
(c) The functions appear to be equal: fit) = git) = hit)
Analytically,
/3\2r/3
/W=4| =4
4(^)' = ^(0
;i(t) = 4^-0.653886, = 4|-g-0.653886J, = 4(0.52002)'
^9i/3Y
«(f) = 41
4(0.52002)'
No. The definite integrals over a given interval may be
equal when the functions are not equal.
97.
=
IQQOe-"'^'
dt
Jo
=
\ 2000 „,1
.-0.06
10
0
=s;
$15,039.61
99.
101. False, e is an irrational number.
t
0
1
2
3
4
y
1200
720
432
259.20
155.52
y = Cik!)
When r = 0. y = 1200 => C = 1200.
y = 1200(/f )
720 432 259.20 155.52
1200 - °-^- 720 - °-^' 432 " "■^- 259.20
Let k = 0.6.
y = 1200(0.6)'
103. True.
105. True.
figix))
= 2 + e'-'f^-^
£M = .^and£[e-1=-
= 2 + .r - 2
= X
g{f{x))
= ln(2 + e- -
e' = e'" when .v = 0.
= In e^ = X
ie°){-e-°) = -1
= 0.6
246 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
f=^>(! -)-(«) =^
^'^^=idt^'~
y[{5/4) -y] 25 5
lny-ln(|-y)=|f+C
/(r(vi7^)'^^^B*
g(2/5)t + C = Q g(2/5)(
(5/4) - y
y{0) = 1 =^ C, = 4 =» 4e'^5>'
(5/4) - y
4g(2/5)/5 _ ^\ ^ ^ 2^(2/5), = 4g(2/5);y + y = (4e(2/* + l)y
5e(2/5>' 5 1.25
4^(2/5)t +1 4 + g-0.4/ J + 0.256-°"
Section 5.6 Differential Equations: Growth and Decay
l.^ = . + 2 3. ? = . + 2
abc dx
X?- dy
• >> = I (.X + 2)dx = y + 2x + C jj^ = A
■■ ■ ■•■ )jh'^y = j'^
Injy + 2| = jc + Ci
>- + 2 = e^**^' = Cr'
>> = Ce^ - 2
,_5jc 7. y' = V^ );
W = 5x y
\yy'dx= ISxdx ■- \—dx= iVxdx
lydy=\5xdx . j^^}^'^
i,2 = |,2^C, ln, = f.3/2 + C,
,2 _ 5,3 = c ^ = ^'^^^'"'"'^-
= gC, g(2/3);(5/^
Section 5.6 Differential Equations: Growth and Decay 247
9. (I + x^)y' - 2xy = 0
1 +.t2
2x
y \ + x^
jjdx = jj^^dx
- dx
(dv^ f 2x
J y Ji+x'
Inv = ln(l + X-) + Ci
\ny = ln(l + x^) + In C
Iny = lnC(l + x'^)
y = C(l + x^)
11.
dQ^k
dt t'
dt
dQ= -- + C
t
dN
13. =r- = k{250 - s)
ds
If-/
k{250 - s)ds
dN = -x(250 - 5)2 + C
A' = -t(250 - ip + C
15. (a)
(b)
di
dx
_dy_
y-6
x(6->), (0,0)
n
Inb - 6| = ^- + C
>- - 6 = e--""/2 + c = c,e-'^/2
.V = 6 + C.g-'^'/^
(0, 0): 0 = 6 + Ci =^ Ci = -6 ^ y = 6 - 6e-^/2
17. ^4,. (0.10,
/-/■
rrfr
y = -r^ + C
10 = -(0)2 + c
y = -r2+10
C= 10
19. |=_i,, (0,10)
/f^H
\dt
In V
f + Ci
y = e-(r/2) + C, == gC, g-,/2 = Cg-r/2
10 = Ce° :^ C = 10
y = lOe-'/^
y = Ce'-'' (Theorem 5.16)
(0, 4): 4 = Ce" = C
(3,10): 10 = 4e3* =4. /t = | ln[|
When x = (,,y = Ae^'^ ^^(^nm = 4^^(5/2)^
23. ^ = kV
dt
V=Ce*' (Theorem 5.16)
(0, 20,000): C = 20,000
(4, 12.500): 12,500 = 20.000e* ^ <•' = ^ ln(|
When r = 6, V = 20,000e '-""''■ *^'*' = 20,O0Oe^^5/8)'«
= 20,000(|j " = 9882.118
248 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
25. V = C<^\ ( 0, -1, (5, 5)
^ = 2
, = -^
5 = 2^*
In 10
k =
y = i g(ta 10/5), = l(10'/5} or 3^ = ieO.4605,
29. A differential equation in x and y is an equation that
involves x, y and derivatives oiy.
27.
y = Ce^, (1,1), (5,5)
1 = Ce*
5 = Ce5*
5Ce* = Ce5*
Set = g5«.
5 = 6^"
/t = ^ - 0.4024
4
■y = Ce°-*'2'"
1 = CfiO'W^i
C « 0.6687 (C = 5-1'"')
y - 0.6687e0-^24,
31.
rfy 1
-dx = -2^
dy
— > 0 when xy > 0. Quadrants I and in.
33. Since the initial quantity is 10 grams, y = iOeIi"(i/2)/i62o],_ ■^^hen / = 1000, y = iQ^^iUD/iiiojvxa) « 552 grams. When
/ = 10,000, >' = iOe['"<i/2Vi620](io,ooo) « 0.14 gram.
35. Since >- = c^^^U2)/\(,io],^ ^g ^ave 0.5 = Cet'"('/2Vi620]Cio,ooo) ==^ c = 36.07.
Initial quantity: 36.07 grams.
When t = 1000, we have y = Ce[i"(i/2)/i62o](iooo) ^ 23.51 grams.
37. Since the initial quantity is 5 grams, we have y = 5.0e['"*'''^'^^'^*.
When? = 1000,>' = 4.43 g.
Whenf= 10,000, v=» 1.49 g. ''
39. Since y = Ce^^iUD/i'i.im ^ ^e have 2.1 = Cet"'"/2»/24.360](iooo) ==> c = 2.16. Thus, the initial quantity is 2.16 grams. When
t = 10,000, y = 2.16e['"('/2'/2«*]«'°'0'») = 1.63 grams.
dy
41. Since— = ky. y = Ce'" or y = y^e'".
1
7% = yoe
1620*
-In 2
1620
y = y g-'ln2)//I620_
When t= 100, y = y^e''^^" 2)/i6.2 „ ^^(0.9581).
Therefore, 95.81 % of the present amount still exists.
43. Since A = 1000e°°^', the time to double is given by
2000 = 1000e°°'5' and we have
2 = gO.oer
In 2 = 0.06f
' = o:o6'"-^^y'''''-
Amount after 10 years: A = 1000e*°°«'<'°' = $1822.12
Section 5.6 Differential Equations: Growth and Decay 249
45. Since A = 750e" and A = 1500 when t = 7.75, we have
the following.
1500 = 750s''"'-
r = :^ = 0.0894 = 8.94%
7.75
Amount after 10 years: A = 750eO"89''(io) = $1833.67
47. Since A = 500e" and/4 = 1292.85 when t = 10, we have
the following.
1292.85 = SOOe'O'-
ln(1292.85/500)
'■ = lo
The time to double is given by
1000 = 500e'"»5o,
In 2
= 0.0950 = 9.50%
0.095
= 7.30 years.
0 075\<i2)(20)
49. 500,000 = P| 1 + ^^ j
0.075 Vz-w
P = 500,000 1 +
= $112,087.09
12
51. 500,000 = P 1 +
P = 500,000 1
0^yi2K35)
12
0.08
12
$30,688.87
53. (a) 2000 = 1000(1 + 0.07)
2 = 1.07'
ln2 = r In 1.07
In 2
t =
In 1.07
=» 10.24 years
/ 0.07 V^'
(b) 2000 = 1000( 1 + —1
2=1 +
0.007
12
ln2 = 12fln 1 +
0.07
12
In:
121n(l + (0.07/12))
55. (a) 2000 = 1000(1 + 0.085)'
2 = 1.085'
ln2 = rlnl.085
In 2
9.93 years
In 1.085
= 8.50 years
(b) 2000 = 1000( 1 +
2 = (l +
0.085
12
0.085\'2'
12
ln2 = 12rln 1 +
0.085
12
/ =
In 2
12
1 (i ^0-085\
8.18 years
(c) 2000
2
In 2
f =
(d) 2000
2
In 2
/ 0.07 V*"
0.07 Y«'
365/
3651n(ll"(0.07/365))^^-^°y^^
■- lOOOeCO""'
; gO.cni
' O.Olt
In 2
■■ — = 9.90 years
(c) 2000 = 1000 1 +
0.085 V"'
1 +
365 /
0.085 y*"
365 /
, -, -,.. , 1 , 0.085\
ln2 = 365rln( 1 + "ttt" I
t =
365
(d) 2000
0
Inh +
lOOOfOO*"
,0.085;
ln2 ^ ,^
365
e'
hi 2 = 0.085f
_ ln2
' ~ 0.085
= 8.15 years
250 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
57. P = Ce*' = Ce-oow
P(-l) = 8.2 = Ce-o«"<-" ^ C = 8.1265
P = 8.1265e-o™9'
P( 10) = 7.43 or 7,430,000 people in 2010
59. P = Ce'" = Cgoo^^'
P{-\) = 4.6 = Cg0036(-i) => c = 4.7686
P = 4.7686eO'>36'
/'(lO) = 6.83 or 6,830,000 people in 2010
61. If fc < 0, the population decreases.
If ^ > 0, the population increases.
63. P = Ce^ (0, 760), (1000, 672.71)
C = 760
672.71 = 760e'««^
ln(672.7 1/760) . nmmo-7
^ = loOO -0000122
P = 760e-0'»0'22x
When X = 3000, P = 527.06 mm Hg.
65. (a) 19 = 30(1 - e^"*)
30e20t = 11
, = MM„_0.0502
N^ 30(1 - e-00502,)
(b) 25 = 30(1 - e-oo502» )
„- 0.0502/ = _
6
-In 6
^ = ^0050^"'^'^^^^
67. 5 = Ce*/'
(a) 5 = 5 when f = 1
5 = Ce*
lim Ce'/' = C = 30
r — »oo
5 = 30e*
A: = Ing « -1.791S
5 » 30e-'™i8'''
(b) When f = 5, 5 = 20.9646 which is 20,965 units.
(C) 30
69. Ait) = V(r)e-o'0' = 100,000e°^-^e'-''"" = 100,0006"^-^^°""
^ = 100,OOof^ - O.loV"'^''""" = 0 when 16.
dt \Jt I
The timber should be harvested in the year 2014, (1998 + 16). Note: You could also use a graphing utility to graph A{t)
and find the maximum of A(r). Use the viewing rectangle 0 < x < 30 and Q < y < 600,000.
71. ;8(/)= 101og,of/o= 10
/n
10"
(a) /3(10-"') = lOlogjorrr;^ = 20 decibels
(b) )3(10-9) = lOlogioTT^TTi = 70 decibels
10-
10""
(c) /3(10-") = lOlog.o-y^TT^ = 95 decibels
(d) /SdO-") = lOlog.o-— Y^ = 120 decibels
10"
10-
73. R = '"/.„°. / = e«in 10 = io«
(a) 8.3 =
In 10
In/- 0
In 10
/ = 1083 « 199,526,231.5
^ _ g2Rln 10 — gifln 10 _ /^Slu 10)2 = (10" )2
Increases by a factor of e^ "^ '° or 10'*.
^ dR 1
rf/ / In 10
Section 5.7 Differential Equations: Separation of Variables 251
75. False. \iy = Ce^, y ' = Cke^ i- constant.
77. True
Section 5.7 Differential Equations: Separation of Variables
1, Differential equation: y' = ^y
Solution: y = Ce''"^
Check: y' = ^Ce^'' = 4v
3. Differential equation: y" + v = 0
Solution: >" = C, cos x + Cj sin j:
Check: y' = — Cj sin or + Cj cos x
y" = —C^ cos x — C, sin x
y" + y = — Ci cos X — Cj sin JT + Cj cos j: + C2 sin j: = 0
5. y = — cos -t Injsec x + tan j:
1
y' = (—cos a:)
sec jc + tan ;c
cos X
{secx • tanj: + sec'j:) + sin j: Injsec.x + tanjcj
sec j: + tan ;c
= — 1 + sin JT ln|sec -t + tan.ic
1
secx)(tan.t + secj:) + sin x In I seer + tan a: I
y"= (sinx)
(sec.r • tan.t + sec^x) + cosxln|secx + tan.r|
sect + tan.r
= (sinx)(secjt) + cosxln|secj: + tanx|
Substituting,
y" + y = (sin x) (sec x) + cosxln|secx + tanjc| — cos j: ln|secx + tanx|
= tanx.
In Exercises 7-11, the differential equation is j'''*' — 16y = 0.
y = 3 cos X
yW - i6y = -45C0SJC ^ 0,
No.
y = e
v(4)
Yes.
16y = 16e-^ - \6e-^ = 0,
11. y = Cifi^ + C,)?--"^ + C3 sin 2x + C4 cos 2x
y*"' = 16Cie^ + leCjg-^ + I6C3 sin Ix: + 16Q cos 2x
yW - 16y = 0,
Yes.
252 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
In 13-17, the differential equation is xy ' — 2> = x^e'.
13. y = X'^, y' = 2x
xy' -2y = x{2x) - 2(;c2) = 0 ¥= x^e'
No.
15. y = xH2 + e^), y' = xV) + 2x(2 + e^)
xy' - 2y = x[xV + Ixe' + 4x] - 2[xV + 2x^] = xV,
Yes.
17. y = lnx,3'' = -
xy ' — 2y = X
2 In X ^ x^e^. No.
19. V = Ce*^
dy
dx
= Cke'^
Since dy/dx = O.OTy, we have Cke'^' = O.OTCe*".
Thus, /t = 0.07.
21. y- = Cx^ passes through (4, 4)
16 = C(64) => C = i
Particular solution: y^ = j x^ or 4^^ = x^
23. Differential equation: 4yy ' — x = 0
General solution: 4y~ — x- = C
Particular solutions: C = 0, Two intersecting lines
C = ±1, C = ±4, Hyperbolas
\.
.^
^
^^
^"**^*>^-.
p^
"^^--^
"-H^,^
c = ~\ .^
^v
— • ^^*T' ' '
C = -4
25. Differential equation: y ' + 2>' = 0
General Solution: y = Ce~^
y' + 2y = C(-2)e-^ + 2{Ce-^) = 0
Initial condition: y(0) = 3, 3 = Ce° = C
Particular solution: y = 3e~^
27. Differential equation: y" + 9y = 0
General solution: y = C, sin 3a: + Cj cos 3x
y' = 3C| cos 3x — 3C2 sin 3x,
y" = - 9C, sin 3x - QC, cos 3x
y" + 9y = (-9C, sin 3;c - 9C2 cos 3x) +
9(C, sin 3x + C cos 3x) = 0
Initial conditions: y(— -) = 2, y'( — j = 1
2 = C| sin( ^ 1 + Co cosi —
y ' = 3C] cos 3x — 30, sin 3x
1 = 3C, cos( y 1 - 3C2 sin( -|
C. = 2
= -3C,
C, =
1
Particular solution: y = 2 sin 3x — - cos 3x
Section 5.7 Differential Equations: Separation of Variables 253
29. Differential equation: x^y" - 3xy' + By = 0
General solution: y = C,x + Cjx'
y' = C, + 2C2x'^,y" = 6C2X
xy- 3xy' + 3y = x^iec^x) - 3x(C, + BQjc^) +
3{C,x + C^x^) = 0
Initial conditions: y(2) = 0, y '(2) = 4
0 = 2C, + 8C2
C, + 4C2 = 0
C, + I2C2 = 4
Q =2- Ci = -2
Particular solution: y = — 2jr + — jc^
dx
= /
3x^dx = x^ + C
33. $ =
tic 1 +;c2
(m = 1 + x^, du = 2x dx)
J 1 + .t2
d:t = ^ln(l + x^) + C
35.^ = ^^^=1-2
or j: X
/[-!]■
y = I 1 --Idlx
jc - 2 InUI +C = x-\nx~ + C
dy
31. -r- = sin 2j:
(tc
^/^
y = I sin 2x (ic = — - cos 2x + C
{u = 2x,du = 2dx)
dy
39. -;- = xjx - 3 Let u = Vj: - 3, then x = u- + 3 and dx = 2u du.
dx
xjx -3dx= («2
= uy
+ 3)(M)(2M)rfM
2 Km" + 3M2)rf„ = 2^ J + M^j + C = I {x - 3)'/- + 2(x - 3y'- + C
Al.^ = xe^
dx
43.
-'[
xe^dx = -e^ + C
(m = x'^, du = 2xd!.t)
dy x
dx y
lydy = Ixdx
2 2 ^'
y2-;c2 = C
45. ^ = O.OSr
as
j^ = j 0.05 ds
In |r| = 0.05i + Q
47. (2 +:t)y' = 3y
i y J 2 + .r
(ir
In y = 3 ln{2 + .t) + In C = In C(2 + .r)^
y = C(x + 2)3
254 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
49. yy' = sin jc
= sin ;c dx:
v ^ = s
=T- = - cos JC + C,
y^ = -Icosx + C
dy
51. yr^"4? ^ = X
dx
dy =
VT-4?
dx
\-'U
4x^
dx
4/<.-
4x^)-'^H-Sxdx)
y = -i(l - 4;c2)'/2 + C
53. _y In j: — xy' = 0
lf = /
— (fa \ u = In X, du = —
X \ X
liiy = -(ln;c)2 + C,
y = g(l/2)(bu:F + C, = (^g(lnx)V2
55. yy' - e^ = 0
jydy = j,
e^dx
y2 = 2e^ + C
Initial condition: yiO) = 4, 16 = 2 + C, C = 14
Particular solution: y^ = le' + 14
57. >'(x + I) + y' = 0
/?-/'
Iny = + C,
Initial condition: ^(-l) = 1, 1 = Ce-'^^ C = e^^^
Particular solution: v = e[i-(A:+i«/2 = g-(.r^+2r)/2
59. ^(1 +x^)^ = xil +y^)
^ dy = T— — ^ dx
1 +>'-•' l+x^
iln(l +y2) = ^ln(l +;t2) + C,
ln(l + y2) = ln(l + x^) + In C = ln[C(l + x^)]
\ +y^= C{1 + x^)
.v(0) =73:1+3 = C^C = 4
1 + 3,2 = 4(1 + ;r2)
;y2 = 3 + 4;t2
61. — - = Mv sm V-
|3V
— = V sin v^
J " J
dv
Ihm = —-cos v^ + Ci
Initial condition: u(Q) = 1, C =
-1/2
= ^1/2
63. dP - kPdt = 0
/f = f
In P = to + C,
Initial condition: ^(0) = Pg, Pg = Ce° = C
Particular solution; P = P^e'^
Particular solution: u = e(i-':osv^)/2
Section 5.7 Differential Equations: Separation of Variables 255
65.
dy _ -9x
dx " 16y
\\6ydy= - \9xdx
By
-9
2 = -^x2 + C
9 25
Initial condition: y(l)= 1, 8= -- + C, C = —
-9 25
Particular solution: 8y- = ^t-jc^ + ^r.
•^2 2
16y2 + 9x^ = 25
67. m
_ dy 0 -y
/f=H
£tc (x + 2) - X
Iny = -2"^ + C,
y = Ce-^/^
69. /(.r,^) =;c3 - 4x>'2 + y3
f(tx,ty) = i'.c^ - 4rjcf'y^ + t^ y^
= r^U^ - 4x>'2 + y)
Homogeneous of degree 3
71. f{x,y) =
fitx, ty) --
xY
fx'Y
= t=-
xY
Homogeneous of degree 3
73.
f{x. y) = 2 In
/(rx, /y) = 2 In
xy
txty
= 21nr2j:j, =
2(ln f '
+ In
xy)
Not homogeneous
77.
>-' =
x + y
2x '
y = vx
V + X— =
dx
X + vx
Ix
•
dv
1 + V
2
- V
J dv ^
1 - V
'dx
X
-ln(l - v)2 = ln|;c| + In C = ln|Cx|
1
(1 - v2)
1
\Cx\
[1 - (y/x)]
X
2 = \CX\
= \Cx\
(x - y?
\x\ = C(x - y)2
75. f(x, y) = 2 In =
/(a,ry) = 21n— = 21n-
ty y
Homogeneous degree 0
79.
X - y
V = vx
y
X
+
y '
V +
dv
'Tx
=
X
X
+
XV
XV
V
dx +
rdv
^
\_
-
^dx
1 + V
f v+1 , {dx
-ln|v2 + 2v - 1| = -ln|.v| + In Q = in
C,
c
|v^
+ 2v-
^1=:;^
V-
_ V
c
+ 2- -
1
X-
.T
x-
|y- + 2v>'-.T-| = C
256 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
81.
V + X
y
dv
xy
— ^- y = vx
x-^ — y-
.2 _ ^2„2
dx x^ — xf-v
V dx + x dv
1 - V
-dx
/^-/^
-y-j - ln|v| = \n\x\ + InC, = ln|Ci.r|
-1
2v2
= ln|Ci;cv|
-x^
2y'-
= ln|C,);|
y =
= ce-'^'^y'
83. xdy ~ {Ixe'yl^ + y)dx = Q,y = vx
x{vdx + X dv) - {2x6'" + vxjdx = 0
je^'dv = jl
dx
e'' = In C,;c'
«>■/•' = In C, + In x2
e^A = C + lnx^
Initial condition: y(l) = 0, 1 = C
Particular solution: e^^' = 1 + In x^
85. \x sec- + y]dx - xdy = 0,y = vx
[x sec V + xv)dx — xiv dx + x dv) — 0
(sec v + v)dx = vdx + xdv
f , (^
cos V dv = —
sin V = In X + In C,
Initial condition: ^(l) = 0, 1 = Ce° = C
Particular solution: x = e*'"'^/-^*
87.^ = .
dx
/
A:(ic = -j;- + C
rfy
89.| = 4-y
In \4 - y\ = -X + Ci
4-y
y = 4 + Ce"-'
91. ^ = O.Sy.yCO) = 6
93. ^ = O.Olj'llO - j), y(0)
12
I I I I I I { If t I I I I
I I I I I I \ I I I I I I
I I I I I I Ifi I I / i I
/ y / / y ^ y J ^ / / /
y y y y ^ 1( y y y y y y
y y y y
^ y y y y
'■///////
,'/ ///////
■//////////
^y y y y y y y y y y
Section 5.7 Differential Equations: Separation of Variables 257
95. ^ = ky,y = C^
dt
Initial conditions: >'(0) = y^
>'(1620)
2
C
= ^-0
Jo
2
= ^-ofi'"*
t
ln(l/2)
1620
Particular solution: y = ;yQe-'(iii2)/i62o
When f = 25, v =» 0.989yo. y = 98.9% of yo-
99. ^ = ky(y- 4)
97. ^ = kiv - 4)
(fa
The direction field satisfies (dy/dx) = 0 along > = 4; but
not along y = 0. Matches (a).
The direction field satisfies {dy/dx) = 0 along y = Q and y = A. Matches (c).
101.
^ = /t(1200 - w)
dt
J 1200 - w J
ln|1200 - M'l = -fa + C,
1200 - w = e
_ ^-fa+C,
Ce-
w = 1200 - Ce"*'
^(0) = 60 = 1200 - C =* C = 1200 - 60 = 1 140
w = 1200 - 1140e-*'
(a) 1400
103. (a)
f-«--"
J IV - V J
-ln|W- v| = /tr + C,
V = W - Ce^*'
Initial conditions:
W = 20, V = 0 when ; = 0. and
V = 5 when r = 1 .
C = 20,k= - ln(3/4)
Particular solution:
V = 20(1 - ei"'3/4).) = 20( 1
(b) k = 0.8: / = 1.31 years
k = 0.9: t = 1.16 years
k= 1.0: t= 1.05 years
(c) Maximum weight 1200 pounds
V = 20(1 - e-0.2877,)
(b) .? = 120(1 - e-o-28^')(/f
= 20[? + 3.476 l<'-o-S'''
C
Since j(0) = 0. C « - 69.5 and we have
j«20/ + 69.5(e-°-2"'' - 1).
lim w = 1200
/-»oo
258 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
105. Given family (circles): x'^ + y"^ = C
2x + 2yy' = 0
Orthogonal trajectory (lines): y' = '
If'f
dx
x
Iny = \nx + \n K
y = Kx
107. Given family (parabolas): x^ = Cy
2x = Cy'
, _ 2x _ 2x _2y
C x^/y X
Orthogonal trajectory (ellipses):
\
4(
p
®
^y
lly dy = — \xdx
x^ + 2y^ = K
109. Given family: y'^ = Cx^
2yy' = 3Cx^
, _ 3Cx^ _ 3x2 /y2\ _ 3y
2y 2y \x? / 2x
Orthogonal trajectory (ellipses):
y =
2x
3Jydy^-2Jx
-x^ + Ki
3\y dy — —2\xdx
3y2
33,2 + 2x2 = K
111. A general solution of order n has n arbitrary constants
while in a particular solution initial conditions are given
in order to solve for all these constants.
113. M{x, y)dx + N{x, y)dy = 0, where M and A' are
homogeneous functions of the same degree.
115. False. Consider Example 2. v = ^c^ is a solution to
xy ' - 3y = Q,h\Ay = x^ + Ws not a solution.
117. False
f(tx,ty) = tV + t^-xy + 2
*t2f(x,y)
Section 5.8 Inverse Trigonometric Functions: Differentiation 259
Section 5.8 Inverse Trigonometric Functions: Differentiation
1. >> = arcsinjt
(a)
X
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
y
-1.571
-0.927
-0.644
-0.412
-0.201
0
0.201
0.412
0.644
0.927
1.571
(c)
(d) Symmetric about origin:
arcsin(— or) = — arcsinx
Intercept: (0, 0)
3. False.
1 TT
arccos - = T
since the range is [0, tt].
. . 1 77
5. arcsin - = —
2 6
1 -n-
7. arccos t^ = v
9. arctan
73 ^ TT
3 6
11. arccsc(-72)
4
13. arccos(-0.8) « 2.50
15. arcsec(1.269) = arccos 7-77-
\ 1.269
= 0.66
17. (a) sin( arctan 4) = 5
(b) sec( arcsin 5
5
5; ~ 3
19. (a) cot
-^
(b) CSC arctan! - —
11
5
21. y = cos(arcsin 2t)
6 = arcsin 2x
y = cos 0 = J\ - 4a-
23. V = sin(arcsec .r)
6 = arcsec x,Q < 6 < n. 6 i^
y = sinO
The absolute value bars on x are necessary because of
the restriction 0 < 6 < tt. 6 ^ tt/1. and sin 6 for this
domain must always be nonnegative.
260 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
25. y = tanf ;
X
arcsec -
0 = arcsec -
y = tan 6
Jx^'^
27. y = cscl arctan
X
72
6 = arctan
y = CSC I
29. sin(arctan 2r)
2x
yrT4?
/=«
Asymptotes: y = ±\
arctan lie = 0
tan e = Ir
sin e =
2x
vT+4j?
VT+T?
31. arcsin(3x — tt) = 5
3jc — 77 = sin^j)
X = 5[sin(5) + 77] = 1.207
33. arcsinV2x = arccos Jx
v^ = sin(arccos ~Jx)
Jlx = Vl -^ 0 < a: < 1
2x= 1 -;c
3a: = 1
1
^ = 3
35. (a) arccscx = arcsin- |jc| > 1
Let y = arccsc x. Then for
77 77
-— < y < 0 and Q < y < —,
csc>' = ;c => siny = \/x. Thus, y = arcsin(l/j:)
Therefore, arccsc j: = arcsin(l/j:).
1 77
(b) arctan x + arctan - = — , x > 0
X 2
Lety = arctan X + arctan(l/jr). Then,
tan(arctanx) + tan[arctan(l/x)]
tan V
1 - tan(arctanx) tan[arctan(l/x)]
X + (lA)
1 - x(iA)
X + (1/x)
0
(which is undefined).
Thus, >» = 77/2. Therefore, arctan x + arctan(l/x) = 77/2.
37. f(x) = arcsin(x — 1)
X — 1 = sin>'
X = 1 + siny
Domain: [0,2]
Range:
77 77
'2' 2
/(x) is the graph of arcsin x
shifted 1 unit to the right.
n-
-
f-
• /(2-f)
-1
-It'
■("•-f)
39. fix) = arcsec 2x
2x = secy
1
X = — sec y
Domain: (-°°. -|]. [|. '
Range: [0,f),(f,77]
^i-ol
Section 5.8 Inverse Trigonometric Functions: Differentiation 261
41. f(x) = 2 arcsinU - 1)
fix) = ^
Vl - (jt - If V2x-x^
43. g{x) = 3 arccos —
g'ix) =
-3(1/2)
Vl - U-V4) V^^^
45. /(jc) = arctan '
47. g(x)
arcsin 3x
Ma
•^'^""^ 1 + UVa") a
+ x2
g'W
x\hl J\ - ^yr) - arcsin 'hx
3;c - yi - 9j:- arcsin 3.t
49. /!(/) = sin(arccos t) = Vl - t-
//'«=|(l-^2)-'/^(-2f) = -==
2 Vl - r
51. y = X arccos j: - Vl - -t-
>> = arccos X —
VT^
(1 -.r2)-i/2(-2A:)
= arccos x
53. y = - - In + arctan x
2\2 j: - 1
= T[ln(jr + 1) - ln(.t - 1)] + - arctan jc
55. y = X arcsin x + Vl — x-
dx "VVT
+ arcsin A-
VT^
dy _ 1/ 1
1 \ ^ 1/2
dx 4\j; + 1 j: - 1 / 1 + x^ I - x^
1
57. y = 8 arcsin
x ArVl6 — x^
59. J = arctan .t +
1 +x'-
y'=2-
1^ - ^^\ ■'' - f(16 - x^)-/2(-2x)
Vl - U/4)- 2 4
8
Vl6^^
Vl6 - x2
16 - (16 - x^) + x^
2V16 - X-
X-
2V16 - X- Vl6 - x2
3' =
1 (1 + X-) - x(lt)
1 + x^ (1 + x^y-
(1 + X-) + (1 - x^)
(1 + x=)=
(1 + x^r
61. /(x) = arcsin x, a = -
fix)
yr
fix) =
(1 - x2)3/2
.,,,., (l).,.(i)(,_i).
1\ 77 , 2V3/ 1
6^ 3
""2
^2W=/U +/'!
1\ . 1 ,,/l
-^f/lfH-
I\- TT , 2v/3/ 1
^¥('4J
262 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
63. fix) = arcsec x - x
1
fix) =
\x\V^^^l
= Owhen \x\^x'^ - 1 = 1.
jc2(;c2 - 1) = 1
X* — x^ — I = 0 when x'^ = or
/i + Vs
Jf = ± W 2 " ±1-272
Relative maximum: (1.272,-0.606)
Relative minimum: (- 1.272, 3.747)
65. f{x) = arctanx — arctan(j: — 4)
\ + x^ 1 + U - 4)^
1 + .^2 = 1 + (;c - 4)2
0 = -8jc + 16
x = 2
By the First Derivative Test, (2, 2.214) is a relative
maximum.
67. The trigonometric functions are not one-to-one on
(— oo, oo), sot their domains must be restricted to
intervals on which they are one-to-one.
69. y = arccot x,Q < y < tt
X = cot V
1
tany = -
X
So, graph the function
y = arctani - J for ;c > 0 and y = arctan
©-
ioxx < 0.
\
71. (a) cot 0 = -
6 = arccoti -
(b)
de
dt
5 dx
-5 dx
+ (l\ ^' ^' "•" 25 A
If ^ = -400 and a: = 10, ^ = 16 rad/hr.
If ^ = -400 and ;c = 3, ^ = 58.824 rad/hr.
dt dt
73. (a) h(t) = -\(,t' + 256
- 161^ -I- 256 = 0 when t = 4 sec.
h -16f2 4-256
(b) tane
500
0 = arctan
de
500
-8t/125
•lOOOf
dt 1 -h [(4/125)(-/2 + 16)P 15,625 -^ 16(16 - t-Y
When f = 1, de/dt « -0.0520 rad/sec.
When t = 2, rf^M « -0.11 16 rad/sec.
Section 5.9 Inverse Trigonometric Functions: Integration 263
, , tan(arctan x) + tan(arctan y) x + y
75. tan(arctan x + arctan y) = -; -, ; -, r = -; , xy i= I
1 — tan(arctan x) tan(arctan y) 1 — xy
Therefore,
arctan x + arctan y = arctanl ],xy¥' 1.
\ 1 - xyj
Let X = 2 3rid y = 5 .
/1\ (A (1/2) + (1/3) 5/6
arctan^-j + arctan^-j = arctan ^ _ ^^j^^) . (1/3)] = ^'^'^
5/6 , TT
l-(l/6) =arctan3^ = arctanl =-
77. f(x) = be + sin.ic
fix) = k + cosx > QfoTk > I
f'(x) = (t + cos ;c < 0 for A: < - 1
Therefore, /(x) = fcc + sin jc is strictly monotonic and has an inverse for k <
loTk > 1.
79. True
81. True
dr -, 1
—-j arctan jcj = ? > 0 for all x.
dx 1 + X-
d^ , ,T sec^;c sec^x
— H arctan(tan x)\ = ;— = — ;— = 1
dx I + tan-x sec-x
Section 5.9 Inverse Trigonometric Functions: Integration
■•/
79^^
dx = 5 arcsin —] + C
Ki)
3. Let u = 3x, du = 3 dx.
dx = ^\ S3) dx
Jo yr=^? 3jo vi - (3x -'
- arcsin(3x)
0 "Ts
Jt^* = I
4 arctanl -I + C
7. Let u = Zx, du = 2 dx.
■V3/2 , , rV3/2
r^'^=_i , 1 [^"' 2 ,
- arctan(2x)
v^/2
— 2L
0 ~ 6
J xV4x2 - 1 J
2xV(2x)= - 1
tZv = arcsec|2x| + C
r x^ fr t1 r if it ii
11. ;' dr = X - -, , oLv- = X civ - - , civ = -x^ - - ln(x^ + 1) + C (Use long division.)
13. ^=a[v = arcsin(x + 1) + C
J VI - (x + 1)2
15. Let i< = t-.du = 2tdt.
J v'l -t' 2j ,
(2r) dt = - arcsin(f =) + C
'1 - (r
264 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
17. Let u = arcsin x, du
yr^
dx.
0 >
'o Vl -X2
dx
0 32
0.308
19. Let «= 1 - x^, du = -Ixdx.
X
\ -j===dx=M {\-x')-'l\-lx)dx
J-I/2VI - X^ ^J-\/2
-vr^
= -0.134
0 _ 73 -2
-1/2 2
21. LetM = e'^,du = le^ dx.
1 r 2e^
j4 + e^- 2J4 +
1 e^
, ,,,, dx = — arctan -r- + C
{e^f 4 2
25.
f \
J v^ -Jl — X
I J ,{2udu) = l\
J MVl - M- J
dLr. u = ^, X — u^, dx = 2u du
du
= 2 arcsin Vx + C
2 arcsin u + C
-■J7?mw-'i
U-3)
:d:x: +
/:
23. Let M = cosj:, £/m = -s'mxdx.
1 + cos^j:
dx
= - arctan(cos x)\ = —
L Jt/2 4
„ fj: - 3 ^ 1 r 2j; ^ . f 1 _,
27. — ; etc = - -z r dx — 3 \ -z — — dx
J .)c^ + 1 2] X- + \ ] X- + \
1
\n(x^ + 1) - 3 arctan;c + C
V9 - (jc - 3)2 J 79 - U - 3)
. /x- 3
:d[»:
= - 79 - (x - 3)2 - 8 arcsin( -^ ] + C
= - J6x - x^ + 8 arcsini - - 1 ) + C
' }qX- - 2x + 2 Jo 1
1
+ {x- \f
dx =
arctan(ji: - 1
'M
33.
Ix^ + lUn"^ = \x^ + 6x1x3'^ - %^ + L+X3'^ = \x^ + 6x1x3"^ " 4^
= InU^ + 6jj + i3| _ 3 arctan
+ C
{.X + 3)2
-dx
,c r 1 . r 1 _, ■ (x + 2\
35. , = dx = — , , = dx = arcsin — - —
J 7-.r2 _ 4;^ J 74 - (x + 2)2 V 2 ;
37. Let M = -x2 - 4x, d« = (-2x - 4) dx.
j^^^==dx = -||(-x2 - 4x)^'/2(-2x - 4)afe
- 7-x2 - 4x + C
"■ I
2a:- 3
dx =
-274x-x2 + arcsinf^^j = 4 - 273 + -| = L059
74 - (x - 2)
■ dx
Section 5.9 Inverse Trigonometric Functions: Integration 265
41. Let M = ;c2 + l,du = 2xdx.
( X , if 2x
Jx' + lx^ + l 2j(x^+ 1)2 + 1
^ = 2 arctan(;c2 + 1) + C
2m du
43. Let M = Ve' - 3. Then u^ + 3 = e', lu du = e' dt, and "-, "" = dr.
M- + 3
= 2m - 2v^ arctan -^ + C = 2Ve' - 3 - 2^3 arctan
V3
/e' - 3
+ C
45. A f)erfect square trinomial is an expression in x with three terms that factor as a perfect square.
Example: x^ + 6x + 9 = (x + 3)-
47. (a)
(c)
J xj\—x
dx = arcsin x + C, m = ;<:
(b)
yr^^
ate = - Vl -^2 + C, M = 1 - x^
: dt cannot be evaluated using the basic integration rules.
I
49. (a) \Jx-\dx = ^x - 1)3/2 + c, M = X - 1
(b) Let M = V.v- - 1. Then .r = m^ + i and dx = 2m du.
IxV^rn' (& = I (m2 + 1)(m)(2«) dM = 2 I (m-* + u-
)dM = 2(y + j) + C
= :^«^(3«2 + 5) + C = Y^x- l)V2[3(x - 1) + 5] + C = :^(x - l)3/2(3x + 2) + C
(c) Let M = Vx - 1. Then x = m^ + 1 and dx = 2m dM.
|-;r4=^dx = j^^^-i-^(2«) dM = 2 j (m2 + 1) dM = 2[y + M j + C = !«("- + 3) + C = jVx - l{x + 2) + C
Note: In (b) and (c), substitution was necessary before the basic integration rules could be used.
51. (a)
(0,0)
dv
dr \ +T'
y
(0,0)
-/Tf;^^
(0, 0): 0 = 3 arctan(O) + C
y = 3 arctan x
3 arctan x + C
C = 0
3z
2
,.v(3) = 0
55.
= [^ arctanf^^^^ j = ^ arctan(l) = y « 0.3927
266 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
57. Area « (1)(1) = 1
Matches (c)
59. (a) I ^ dtc = 4 arctan x = 4 arctan 1 - 4 arctan 0 = 4( — J -
Let n =
4(0)
(b) Let n = 6.
:dx ~ 4
4 2
+
1 + (1/36) 1 + (1/9) 1 + (1/4) 1 + (4/9) 1 + (25/36) 2
h
3.1415918
(c) 3.1415927
61. (a) —I arcsin - + C ^^, ,
\aj J Vl - {u^/a~)\al Ja^ - u^
1. (a) f [arcsinQ + c] =
[___du__ ^
(b)
Thus^
arcsini - 1 + C.
i " 4- /"I - If u'/a 1 _ 1 r u'
a ^""^a \~ aV\+ {u/a)A ' ai{a^- + u^)/a^i
^, \ du f m' , 1 u ^
Thus, — T = -^ ^ dx = - arctan - + C.
J a~ + u J a'^ + w^ a a
a^ + u^
(c) Assume h > 0.
d_
dx
1 u 1
1
- arcsec — \- C
a a ]
a
u'/a
{u/aU(u/af - 1
™,. r 1^" r «' ,1 i«i „
Thus, — , = — , dx = - arsec -"-J- + C.
J mV«2 - a^ J uju^ - a^ a a
i] =
u^ur — a^)/a^ u^w- — a^
The case « < 0 is handled in a simi-
lar manner.
63. (a) v(t) = -32f + 500
(b) 5(f)
= ^v{t)dt = ^
(-32r+ 500) d;
= - 16f= + 500f + C
5(0) = - 16(0) + 500(0) + C = 0^ C = 0
5(f) = -16f- + 500f
When the object reaches its maximum height,
v(f) = 0.
v(f) = -32f + 500 = 0
-32f = -500
f = 15.625
5(15.625) = - 16(15.625)2 + 500(15.625)
= 3906.25 ft (Maximum height)
—CONTINUED-
Section 5.10 Hyperbolic Functions 267
63. — CONTINUED-
(c)
\l^-'"=-\
kv
dt
Vi2k \V 32 ' '
arctani -. / ^ v'
/32<:f + C
V = tan(c - Vyzici
^ tan(C- V32kt)
k
When r = 0, V = 500, C = arctan(500VA:/32), and we
have
(d) When k = 0.001, v(r) = 732,000 tan[arctan(500V0.00003125) - VO.032 t\
/^ /32
v(f) = ^ / -- tan
k
arctani 500,
v(r) = 0 when t^ = 6.86 sec.
/■6.86 '
(e) /! = 732,000 tan[arctan(500V0.00003125) - 70.032 r] dt
Jo
Simpson's Rule: n = 10; /j « 1088 feet
(f) Air resistance lowers the maximum height.
Section 5.10 Hyperbolic Functions
1. (a) sinh 3 = -
(b) tanh(-2)
10.018
2
sinh(-2) _ e~- - e-
cosh(— 2) e^' + e-
-0.964
5. (a) cosh- '(2) = ln(2 + v^s) - 1.317
04/9)
2/3 /
(b)sech-.(|) = ln(J^4P^U 0.962
3. (a) csch(ln2)
(b) coth(ln5)
2
2 4
^ln2 _ ^-lii2
2 - (1/2) 3
cosh(ln 5)
sinh(ln 5)
gins + g-lnS
^ln5 _ g-]n5
=
5 + (1/5) 13
5 - (1/5) 12
/e^ — e~'\- I ''
7. tanh^ j: + sech- x = I ^ — -r 1 +
e' + g-
£■' + e'
2 ^ir _ 2 + e"^ + 4 e^ + 2 + e'-"
(e' + e-'Y e^ + 2 + e~^
268 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
e' - e^^Xfey + e'A (e^ + e~^\( e^ - e"^
9. sinh X cosh y + cosh x sinh y = \ z II : 1 +
4
= ^{e^y - e-(^+>')] = Y = sinh(A: + .v)
11. 3 sinh a: + 4 sinh' .t = sinh;c(3 + 4sinh^j:) =
2
gX — g-x
3+41
e' — e'
T]
,[3 + e^^ - 2 + e"^] = -(e^ - e-')(e^ + e'^ + 1)
-[e'Jt + e'" + e^ - e' - e'^" - e'"]
sinh(3;c)
13.
sinh a: =
cosh^ ^ ~ ( o ) ~ 1 =^ cosh^ jc = — => cosh x =
3/2 3V13
tanh;ic = —;= — = ,,
713/2 13
CSCh. -3/2 -3
1 2jU
sechjc = — ^ — = —rr—
1 ^
coth X - , , — - ,
3/713 3
15. y =
sinhd - x^)
y' =
-2j:cosh(l - x~)
17. f{x) = ln(sinh;c)
/'W = ^-r(cosh;c) = cothx
sinh
19. y = In tanh
'/2 sech^f^
tanh(V2) \2
1
2 sinh(x/2) cosh(./2)
1
sinh x
= csch;c
21. h{x) = I sinh(2;c) - |
■ ,/ \ 1 , /„ \ 1 cosh(2j:) - 1 . , ,
h'{x) = -cosh(2A:) - - = ^-r^ = sinh^j:
23. fit) = arctan(sinh t)
1
f'(t) =
1 + sinh^ t
cosh t
(cosh t)
cosh' r
= sech /
25. hsty = g{x).
y ^ ycosh X
\ny = cosh a: In j:
\(dy\ _ coshj:
y\dx
+ sinh ;c In ;c
-- = -[cosh X + ;c(sinh x) In x\
ax X
■{cosh X + x(sinh x) In x]
Section 5.10 Hyperbolic Functions 269
27. y = (cosh x - sinh x)^
y ' = 2(cosh X — sinh x)(sinh x — cosh x)
= — 2(cosh X — sinh xY = — 2e"^
29. /(x) = sin X sinh x — cos x cosh x, — 4 < j: < 4
/'(jc) = sin x cosh jc + cos x sinh x — cos x sinh x + sin x cosh jc
= 2 sin x cosh x = 0 when x = 0, ± tt.
Relative maxima: (± tt, cosh tt)
Relative minimum: (0,-1)
f-ff, coshff) i5f;r. coshff)
i — i
31. g(x) = .r sech x =
coshx
(1.20.0.66)
>^
(-1.20.-0.66) -1
Relative maximum: (1.20,0.66)
Relative minimum: (- 1.20, -0.66)
33. y = a sinh x
y' = a cosh x
y" = a sinhx
y '" = a cosh x
Therefore, y'" — y' = 0.
35. /(x) = tanh x
/'(x) = sech-x
/(I) = tanh(l) = 0.7616
1
/'(I) =
cosh-(l)
fU) = -2 sech^x • tanhx /tl) == -0.6397
/',(^) =/(l) +/'(1)(a- - 1) = 0.7616 + 0.42(x - 1)
0.6397
- 0.4200
P.^{x) = 0.7616 + 0.42(x - 1)
-{x - \r-
p./
37. (a) y =10+15 cosh— -15 < x < 15
20 ■•
10
(b) Atx = ±15. y = 10 + 15cosh(l) = 33.146.
At X = 0,y =10+15 cosh(l) = 25.
(c) y' = sinh -j-r. Atx = 15, y' = sinh(l) = 1.175
39. Let M = I - Zx, du = -2dx.
sinh(l - 2x) dx = -^ sinh(l - 2r)(-2) dx
I'
--cosh(l - 2x) + C
41. Let u = cosh(x — 1), du = sinh(x — 1) dx.
i
cosh'(x - 1) sinh(x - Otiv = -cosh^(x - 1) + C
270 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
43. Let u = sinh x, du = cosh x dx.
Tcosh .V
45. Let u = —,du = xdx.
iinhx
■dx = In sinh. V + C
P-f-l
csch^ — \xdx =
-cothy + C
47. Let u — —. du = — rctc.
x x'^
\
csch(l/A-)coth(l/A-)
dx
■[
csch - coth - — x]dx = cscb- + C
X x\ x^ x
49.
25 -A-^ 10 5 - A~" 1015 +,t'
dx + -—\ dx
i"
5 + X
5 - X
o^m'"^ = 5'"^
51. Let u = 2x, du = 2 dx.
r^/2/A
Jo
-dx
rV2/4
Jo
VI - Ax^ Jo VI - [2xf
=(2) dx ■
arcsin(2x)
0 4
53. Let » = x^, du = 2x dx.
J?TT* = i/:
2a , 1 / ■,\ ^
- dx = - arctan(A^j + C
(a^)2 +1 2
55. y = cosh '(3a)
3
^ V9a2 - 1
57. y = sinh~'(tanA)
y' = , (sec^x) = |secA|
Vtan-^ A + 1
59. y = tanh-'(sm2A)
1
1 — sin^ 2a
(2 cos 2a) = 2 sec 2a
61. y = 2Asinh-'(2A) - Vl + 4x^
2
Vl + 4x-
4x
y' = 2a- z + 2 sinh" '(2a) ^=^ = 2 sinh" '(2a)
Vl + 4a2
63. See page 395.
65. y = a sech ' - — Va^ — a-, a > 0
a
-1
^^_
dx (x/a)Vl - (aVq^) ■ Vfl" - a2 aVg^ - A^ ' Va^ - a^ aVq^ - a^
"" + ^
a2 - ^2 - V^5^^
67.
:d^ =
Vl + e^ J e'Vl + (e>p
e^
:a[x = -csch-'(e') + C = -Inl
1 + Vl +
+ C
69. Let iY = Va, du = — tt dx.
2jx
/:
:^A= 2
v^v^l^^ J Vl + (v^)^V2^
— ^) rfA = 2 sinh ' v/J: + C = 2 ln(^ + Vl + a) + C
2Va/
^^•/i^^^ = J(7
-^jr^^'^ = 4ln
(a - 2) - 2
(a - 2) + 2
= 5'"
A- 4
+ C
Section 5.10 Hyperbolic Functions 271
^^' Jl-J-2x^'^'^ - J 3 - 2(1 + l)^'^ - Jj[72U + \)Y - (73)^"^
-1
2^6
In
^{x + 1) - 73
v^U +0 + 73
+ c
276
In
v/2(;c + 1) + 73
72(jt + 1) - 73
+ C
75. Let M = 4jc - 1, ^k = 4 <&.
J 780 + 8^ - ldt= 4J
4 , 1 . /4;c - 1\ ^
, dx = - arcsin — - — + C
781 - {Ax - If 4 \ 9 '
''' y = /sT^'^ = /(- - ^ ^ jrf^)^ = /(- -'^'^^ -«/f^17^
ix
^' . -.20,
— - 4j: + -^ In
2 6
U - 2) + 3
U - 2) - 3
^C=-f-4. + fln
X + 1
X- 5
^ -X . 0,
+ C = — 4.V - — n
2 3
X - 5
.x+ 1
+ C
79. A = 2 sech^atc
Jo •^
81. A
= 2
^/2 + g-xn
dx
Jo ie^^'r +
-dx
8 arctan(e^/2)
8 arctan(e-) - 27r == 5.207
2jo
:dv
7l?FTT
dx
|ln(x-+ ./^^TT)
= - ln(4 + 717) = 5.237
83. l^dt = -^ — dx
J 16 ]x- - lit + 32
3fa^ r
16 J
(^_-^r3^<i^ = ^'"
(.V - 6) - 2
(.V - 6) + 2
+ C = - In
4
.T - 8
X - 4
+ C
When .t = 0: f = 0
C= --ln(2)
When;c = 1:
When t = 20:
t= 10
30A- _ 1
16 ~ 4
In
-3
-4'"(2)=-ln(-
^^ = ^'"6
a(^)-eh4'"^
49 .V - 8
36 2x - 8
eix= 104
104 52 , ^^^ ,
a: = — = -« 1.677 kg
272 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
85. As k increases, the time required for the object to reach the ground increases.
87. y = coshx •
e^ — e~
e^ + e'"
2
= sinhjc
89. y = cosh ' x
cosh y = X
(sinh3')(y') = 1
1
y =:?
smhy Vcosh^ y - 1 ^x^ - 1
91. y = sechx
e" + e^-'
y' = -2(e'^ + e"^)-V - e'")
e^ — e
e' + e ^l\e' + e "
— sech X tanh x
Review Exercises for Chapter 5
1. /(x) = ln;c + 3
Vertical shift 3 units upward
Vertical asymptote: x = Q
12 3 4 5
3- 'n V£^ = 1'"^^" J/+V^' = if'"^^ " ^^ ■" '"^2" + ^^ - '"(^"' + ^^^
5. In 3 + ^ ln(4 - x^) - In jc = In 3 + In 3/4 - x^ - Inx = ln( ^^^
3 V X
7. InVx + 1 = 2
Vjc+ 1 = e2
X + 1 = e"*
X - e" - 1 ^ 53.598
9. g(x) = InVx =2'"^
^'«=t
11. /w = xTinI
1 /; — 1 + 21nx
— j= + V In X = 7=^
2vlnx 2vlnx
13. >- = 75| ln(a + bx) +
^ ^ ±r^ <L
dx if\_a + bx (a +
a + bx
ab
to)" J
(a + bxY
15. y
dy
a \ X / a
1/ fc
a[)c aVa + bx xj x{a + bx)
1
17. u — Ix — 2,du = Idx
1
J 7x - 2 7j
Ix-T ' 7 ' '
Review Exercises for Chapter 5 TTi
, „ f sin .t , r ~ s
19. dx= -\
J 1 + cos j; J 1 +
-dx
1 + cos X J 1 + COS X
-ln|l + cosjtl + C
-r^-K-i)-
X + Inlxl
3 + ln4
23. I secede =
Jo
ln|sec e + tan e|
t/3
= ln(2 + 73)
25. (a) /W = it - 3
1 -,
> = 2-t - 3
2(y + 3)=x .
2(x + 3)=y
r\x) = 2x + 6
(b)
'/
^
>
■^
(c) /-'(/W) =/-'(it - 3) = l{{x - 3) + 6 = ;c
/(/-'W) =/(2;c + 6) = ^(2x + 6) - 3 = j:
27. (a) /U) = v'TTT
y= Jx+ 1
y — 1=0:
x- — \ = y
f-'(x) =x^- \,x > 0
(b)
^
^
■'
(c) /-'(/W) =/-'(7^TT) = VCr - 1)= - 1 = X
■ /(/-W) = /(.t^ - 1) = J{x- - 1) + 1
= v^? = JT for -V > 0.
29. (a) /(x) = 3/.V + 1
V =Vx+ 1
' 3^- 1 =x
x' - 1 =y
/->W=x3- 1
(b)
f-
"^^
y
(c) /-(/W) =/-'(4/^^^) = (^^n^)' - 1 = -r
/(/-'U)) = /(x^ - 1) = ^(.T^ - 1) + 1 = X
31. /U) =y? ^-1
r\x) = U - 2)>/3
(/-')'W = |u-2)-V3
33.
1
35/3
= 0.160
/(.r) = tan .r
/^\ ^ 73
-^Uy 3
f\x) = sec-.r
^U/ 3
3
^ ' \ 3 y /'(7r/6)
4
274 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
35. (a) f{x) = In v^ (b)
y = ln>/t
ey = v^
e'^y = X
e^ = y
/-'W = e^
V
/'V
^
(c) /''(/W) =/"'(ln >A) = e^i"-^ = giBA: = jf
/(/~'W) = Ae^) = In 7^ = In e^ = X
37. y = e-^/2
39. f{x) = ln(e-^) = -jc^
fix) = -lx
41. g(f) = i^e'
g\x) = t-e' + 2te' = te'(t + 2)
43. y = Ve^' + e"^
y' = f(e^ + e-^-)-/2(2e^ - 2e-^) = ' ^ 2.
I ^ e" + e '^
45. 8ix)=~
g'ix)
e^{2x) - JcV x{2 - x)
49. Let M = — 3x^, ^m = —6x dx.
J^-3.^=_i|
£fa:= -7 e-3j:^(-6x)dx= -re"'^ + C
47. y(ln x) + y^ = 0
dy_
(ic x(2y + In x)
51. ^ ^^ Sfa = (e3^-e^ +
e-^)dx
g3jr _ ^ _ g AT ^ ^
3e^ - 3
3e^
+ C
/„-*.-ip.-(-2.,^
53. xe'-^atc
= -/'^ + C
55. Let u = e^ - l,du = e' dx.
e^
1
e^- 1
dx = Ine' - 1 + C
57. y = Wa cos 3x + b sin 3jc)
y ' = e't- 3a sin 3x + 3b cos 3j:) + e*(a cos 3j: + fo sin 3x)
= e^{-3a + b) sin 3x + {a + 3b) cos Sx]
y" = K3(- 3a + b) cos 3jc - 3(a + 3b) sin 3x] + e'[(-3a + b) sin 3x + (a + 3b) cos 3x]
= s^[(-6a - %b) sin 3x + (-8a + 6fe) cos 3x\
y" - 2y ' + lOy = e^{[(-6a - 8fc) - 2(-3a + b) + \Qb] sin 3x + [(-8a + (,b) - 2{a + 3b) + 10a] cos 3a;} = 0
Review Exercises for Chapter 5 275
59. Area
= j xe-'^dx = [-f^'^] = -!(«•"'* - 1) = 0.500
61. y = 33/2
63. y = logjCj: - 1)
65. fix) = 3^-'
/'U) = 3--'ln3
69. g{x) = logjVl - X = - log3(l - x)
g'ix) = ^
1
1
2(1 -;c)ln3 2{x - l)ln3
67. y = x2^-''
Iny = (2x + \)\nx
y' _ Ix + 1
+ l\r\x
y *
(Zx+ 1
3' = >'
+ 2 In .X 1 = ;c^* '( — + 2 In .t
X
71. \{x + l)5(-+»'rfr = 1 As'-*"' + C
J 2 In 5
73. (a) y = x"
y = ax"
75. 10,000 = PefooTXis)
„ 10,000
(b) y = a-
y ' = (In a)<3-'
$3499.38
(c) y = x^ (d) J = a"
In y = JT In ,r y ' = 0
— v' = X • — I- (1) In.v
y- j:
y' = y(l + In.v)
y' = x'(l + In.r)
77. P(/i) = 30e^
P(18,000) = 30e'««»'-- = 15
, _ ln(l/2) _ -ln2
18,000 18,000
Pih) = 30e-<'"° ->/'«•«»
/'(35,000) = 30e-(35.oooin2)/i8,ooo ^ 7 79 inches
79. P = CeOO'S'
2 = eOO'5'
ln2 = 0.015r
In--
t =
0.015
= 46.21 years
81. J = ^i±l
ttc X
h-~l
x^-Ux
y = y + 3 ln|.r| + C
276 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
83. y' -2xy = 0
dv
dx
2xy
\-dy= llxdx
\n\y\ = x^ + Q
y=Ce^
dy v^ ~t~ V*"
85. -^ = ^- (homogeneous differential equation)
dx 2xy
(x^ + y^) dx - 2xy dy = 0
Let y = vx, dy = x dv + v dx.
(x- + vV) dx - 2x{vx){x dv + vdx) = 0
(x^ + vV - 2x^v^) dx-2x>vdv = 0
(x^ — xhP) dx = 2a^v dv
i\-v'^)dx = 2x dv
(dx^ r_2v_
] X J 1 - v2
dv
ln|jc| = -ln|l - v^l + C, = -ln|l - v^l + In C
C C Cx^
1 =
1
— V-
1 -
- {y/xY
X- -
/
Cx
-y2
or
X
x^
^'"x^
-f
87. y = C^x + C^^
y' = Cj + SCjJC'
y" = 6C2X
x^y" - 3xy' + Sy = X'iec^x) - Zx{C^ + SCjX^) + iCyX = C.x^)
= ec^x^ - 3C,x - QCoX^ + 3CiX + SCjX^ = 0
X = 2,y = 0: 0 = 2C, + 8C2 => Cj = -4C2
x = 2,y' = A: 4 = C, + llCj , - '
■ 4 = (-4C2) + I2C2 = 8C2 =» C2 = |, Ci = -2
y = ~2x + -x^
»9.f(x) = 2 arctan(ar + 3)
Review Exercises for Chapter 5 277
91. (a)
Let 0 = arcsin —
sin 6 =
sinl arcsin -I = sm6 = —.
(b)
Let 6 = arcsin
1
sin e = -
cosi arcsin -I = cos 6 = -r~.
93. > = tan(arcsin j:)
vr
95. y = x arcsec x
X
|x|v^^^^
+ arcsec j:
97. y = .t(arcsin x)^ — 2r + 2 Vl — .r- arcsin ;c
2x arcsin x
vr^
+ (arcsin j:)^ — 2 +
2VTZ
2j:
yr
: arcsin x = (arcsin .t)^
99. Let M = e^, (iu = 2e^ dx.
J-TTT^ '^ = \y^ dx = l/yrrU^^^.-) ^ = \ arctan(.-) + C
101. Let « = X-, du = 2xdx.
103. Let M = 16 + X-. du = 2xdx.
(2r) A = — arcsin x^ + C
jjeT^'^^ljleh^^^^'-'-b^^''
+ x^) + C
105. Let u = arctan t- I. <i" = ": ^ <if •
\2/ 4 + x^
. , arctan '-] + C
4\ 2
\Ai -w ■' + <^
\A/ V m
Since y = 0 when f = 0, you have C = 0. Thus,
^\ V
>- = A sin( ^/ - f
109. >- = It - coshv^x
sinhv'x
,^ = 2-^(sinhV^) = 2-^-^
111. Let u = x^. du = 2x dx.
\ ■'' dx = U , ^ ={2x)dx = \\n{x^ + V?~^) + C
278 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
Problem Solving for Chapter 5
1- tan 01 =
tan e, =
10 -X
f{x) = 0, + ^2 ^ arctan(-J + arctanl _
1 +4 ' 1 +
1
36
(10 - x)
= 0
(10 - xf
6
x2 + 9 (10 - xY + 36
(10 - xf + 36 = 2(x2 + 9)
100 - 20;c + x2 + 36 = 2jc2 + 18
;c2 + 20x - 118 = 0
-20 ± V202 -4(-118)
X = = - 10
a = - 10 + v^TS - 4.7648 f(a) =- 1.4153
e = 77 - (0, + e^) « 1.7263 or 98.9°
Endpoints; a = 0: 0 = 1.0304
a = 10: 0 « 1.2793
Maximum is 1.7263 at a = - 10 + 7211 « 4.7648.
/2T8
3. f{x) = sin(lnj:)
(a) Domain: x > Q or (0, oo)
(b) /(x) = 1 = sin(in x) =* In ;c = y + 2^it.
Two values are jt = e'^''^, g(7r/2)+27r_
(c) f(x) = -1 = sin(lnj:)
Two values are x = e""''^, e^''''^.
ln.t = — + 2^77.
(d) Since the range of the sine function is [— 1, 1],
parts (b) and (c) show that the range of/is [— 1, 1].
(e) f'(x) = -cos(lnx)
fix) = 0 ^ cos(!n x) = 0 =^ \nx = — + ktr =i
fie-"'") = 1
/(I) = 0
/(lO) = 0.7440
• Maximum is 1 atx = e^^^ = 4.8105
(f)
[T
lim f(x) seems to be --. (This in incorrect.)
Jr-»0* 2
(g) For the points x = e'"!'^, g-^Vz^ e"'''/^, . . .
we have f(x) = 1.
For the points x = e"''/^, e"^'"!'^, e~^'"'\ . . .
we have f{x) = - 1 .
That is, as j:— ^O"^, there is an infinite number of
points where f{x) = 1, and an infinite number where
f(x) = — l.Thus lim sin(ln a:) does not exist.
You can verifiy this by graphing f{x) on small
intervals close to the origin.
Problem Solving for Chapter 5 279
, , , Area sector t . t , . t
5. (a) 7—r- = T— => Area sector = r— (tt) = -
Area circle Itt Itt 2
, rcosh t
(b) AreaAOP = -{base)(height) - Vjt - 1 dx
rcosh r
1 ^
'^(f) = - cosh t ■ sinh t
\dx
A '(f) = -[cosh- t + sinh- i\ - Vcosh^ t - 1 sinh «
= rCcosh- t + sinh- 1] - sinh^ t
= —[cosh- t - sinh- t] = -
A(f) = -t+C. But, A(Q) = C = O^C = 0
Thus, A(f) = -f or J = 2 A(/).
7. y = In jr
1
y
1,
>> - fe = -(.r - a)
a
y = —X + b — 1 Tangent line
a
lix = Q,c = b - 1. Thus, b - c = b - {b - \) = \.
11. (a)
^
A
,,1-01
jy-""*'= \dt
-0,01
I ^
"-l
1
y0.01
-O.Olr + C
yO.OI
1
C -
- O.Olr
1
y -
(c
- O.Olr)"*
v(0) = l: 1 =^=^ C=l
Hence, y
1
(1 - O.Olr)"
9. Let u = I + Vx, ^ = u - ].x = u^ - 2u + \,
dx = (2m — 2)Jm.
Area = —p dx = — t—
]-,Jx + x h («-!) + ("- - 2u
h u- - u
"I
+ 1)
■du
2 In
2 In 3 - 2 In 2 = 2 Ini
0.8109
(!)
(b)
|y-»*-'rf>-= r
■^— = kt+C,
— E
1
(C - eitr)'/^
.v(0)=.Vo = ^=>C-4
Hence, v
^r
i -)'•■■
For r— > . V— >cx;,
For T = 100, lim v = oo.
r-»r- ■
280 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
13. Since ^ = /t(v - 20),
dt
/t^^-/
dy= \kdt
\n{y - 20) = kt+ C
>> = Ce*' + 20.
When r = 0, y = 72. Therefore, C = 52.
When f = 1, y = 48. Therefore, 48 = Sle" + 20, e* = (28/52) = (7/13), and A: = ln(7/13).
Thus, .V = 52«ti°*''/'3'> + 20.
When f = 5. >- = 52e5M7/i3) + 20 « 22.35°.
JO
15. (a) — = k^S{L - S)
(b) ^ = In m 5(100 - S)
1 -'- Ce-
: is a solution because
rf25 /4
^ = '"19
5|-f) + (100-S)''^
rfr
dr
^= -L(l + Ce-'')-2(-C/te-*')
dt
LCke-
(1 + Ce-'')2
fM
L
Ul
+ Ce-'^-'
(k\
L
CLe-
1 + Ce-*'
U/ 1 + Ce-
L -
1 + Ce-
k^S{L - S), where A:, = — .
L = 100. Also, S = 10 when f = 0 ^. C = 9. And,
S = 20 when t= \ =* fc = -ln(4/9).
Particular Solution. S
100
J + 9gln(4/9)f
100
1 4. 9g-0.8109
(C) 125
^lng)(100-25)f .
= OwhenS = 50or^ = 0.
dt
Choosing S = 50, we have:
50 = 100—
1 + 9ei"M/9)'
2=1+ 9eln(4/9)r
ln(l/9)
ln(4/9)
= r
t = 2.7 months
(d)
(e) Sales will decrease toward the line S = L.
PART II
CHAPTER P
Preparation for Calculus
Section P.l Graphs and Models 282
Section P.2 Linear Models and Rates of Change 287
Section P.3 Functions and Their Graphs 292
Section P.4 Fitting Models to Data 296
Review Exercises 297
Problem Solving 300
CHAPTER P
Preparation for Calculus
Section P.l Graphs and Models
Solutions to Even-Numbered Exercises
2. y= V9 - x^
x-intercepts: (-3,0), (3,0)
y-intercept: (0, 3)
Matches graph (d)
6. y = 6 — 2x
X
-2
-1
0
1
2
3
4
y
10
8
6
4
2
0
-2
4. y = x' — a:
x-intercepts: (0, 0), (- 1, 0), (1, 0)
y-intercept: (0, 0)
Matches graph (c)
8. y = (x - 3)2
X
0
1
2
3
4
5
6
y
9
4
1
0
1
4
9
10. y
x
-3
-2
-1
0
1
2
3
y
2
1
0
-1
0
1
2
12. y = Jx + 2
X
-2
-1
0
2
7
14
y
0
1
V2
2
3
4
5 10 15 20
282
Section P.l Graphs and Models 283
14.
Xmin = -30
Xmax = 30
Xscl = 5
Ymm = -10
Ymax = 50
Yscl = 5
Note that 3; = 10 when x = Ooxx= 10.
16.
j\
J
(a) (-0.5, >-) = (-0.5,2.47)
(b) {x, -4) = (- 1.65, -4) and (x, -4) = (1, -4)
18. y'^ = :? - Ax
^'-intercept: y^ = (fi - 4(0)
y = 0; (0, 0)
jc-intercepts: 0 = x' — 4;c
0 = x{x- 2){x + 2)
x = 0, ±2; (0, 0), (±2, 0)
20. y={x- 1)7^2 + 1
y-intercept: y = (0 - 1)V0^ + 1
y= -1;(0, -1)
x-intercepts: 0 = {x - 1)V?+T
x= 1;(1,0)
22. y
x^ + Zx
[3x + \Y
y-intercept: y =
0^ + 3(0)
[3(0) + 1?
y = 0; (0, 0)
x^ + 7)x
x-intercepts: 0 = 7; — -—
(3x + 1
0
x{x + 3)
(3x + \y
x = 0, -3;(0, 0), (-3,0)
24. >- = 2x - Jx^ + 1
y-intercept: y = 2(0) - VO^ + 1
>.= -1;(0,-1)
x-intercepts: 0 = 2x — Vx^ + 1
2x= Vx^ + 1
4x2 = x2 + 1
3x2 = 1
x^ =-
3
73
' = #(#»
Note: X = — V3/3 is an extraneous solution.
26. >> = x^ - X
No symmetry with respect to either axis or the origin.
28. Symmetric with respect to the origin since
i-y) = (-x)3 + (-x)
—y = — x^ — X
y = }? + X.
30. Symmetric with respect to the x-axis since
x(-yY = xy2 = _ 10.
32. Symmetric with respect to the origin since
(-x)(-^) - V4 - (-x)- = 0
xy - V4--x2 = 0.
284 Chapter P Preparation for Calculus
34. V
a:2 + 1
since y
is symmetric with respect to the v-axis
i-xY- x^
{-xY + 1 .^2 + r
36. I);] — x = 3 is symmetric with respect to the x-axis
since \—y\ — x = ?>
\y\-x = 3.
38.. = -2+2
Intercepts:
(4,0), (0,2)
Symmetry: none
40. 3; = f;c + 1
Intercepts:
(0,1), (-1,0)
Symmetry: none
42. y = x^ + 3
Intercept: (0, 3)
Symmetry: )'-axis
44. y = 2x2 + X = ^(2;f 4. 1)
Intercepts:
(0,0), (-io)
Symmetry: none
50. X = y2 - 4
Intercepts:
(0,2), (0,-2), (-4,0)
Symmetry: x-axis
(-4, Oj
46. y = x^ - 4x
Intercepts:
(0,0), (2,0), (-2,0)
Symmetry: origin
52. y
10
x^+ 1
Intercepts: (0, 10)
Symmetry: y-wdi
-6-4-2 2 4 6
48. y = V9 -x2
Intercepts:
(-3,0), (3,0), (0,3)
Symmetry: y-axis
Domain: [—3, 3]
54. .V = |6 - x|
Intercepts:
(0, 6), (6, 0)
Symmetry: none
Section p. 1 Graphs and Models 285
56. x'^ + Ay- = '^^>y = ±
Va^I?
Intercepts:
(-2,0),(2,0),(0, -1),(0, 1)
Symmetry: origin and both axes
Domain: [—2, 2]
<-.o^
(0,1)
^
(0,-1)
58. 3;c - 4y2 = 8
Af = -ix
y = ±J\x-2
Intercept: Is- 0/
Symmetry: jc-axis
(y^^
'^~^--
60. y - (x + jjU - 2)(j: - j) (other answers possible)
2;c- 13
AX jy —
Lj=f y
3
5x + 3y =
l=>y =
1 -
3
i£
2x- n
1 - 5x
3
3
2x- 13 =
1 - 5x
7.t =
14
X =
2
The corresponding y-
value
isy =
Point of intersection:
(2,-
-3)
-3.
62.
Some possible equations:
x = y^
x=\y\
;c = / + 1
jc^ + >^ = 25
66.
5x - ey = 9 =>
_5x
- 9
6
-7a: + 3y = -18
^y =
7a:-
3
J8
5a: - 9 Ta -
18
6 3
5x - 9 = 14.r -
- 36
27 = 9a:
a: = 3 •
The corresponding v- value is y = 1.
Point of intersection: (3,1)
68. x = 3-f=>f = 3-x
y = X — 1
3 - A = U - 1)2
3 - A = A- - 2i: + 1
0 = .r^ - A - 2 = (x + l)(x - 2)
X = — 1 or X = 2
The corresponding y- values are y = - 2 and y = 1 .
Points of intersection: (- 1, -2), (2, 1)
70. x2 + y2 = 25 ^ >- = 25 - .r=
2x + y = 10=s.y = 10 - 2x
25 - x= = (10 - 2r)=
25 - X- = 100 - 40x + 4x-
0 = 5.v^ - 40x + 75 = 5(x - 3)(x - 5)
X = 3 or X = 5
The corresponding y-values are y = 4 and y = 0.
Points of intersection: (3. 4). (5, 0)
286 Chapter P Preparation for Calculus
72, y = ^ - 4x
y=-{x + 2)
)? - Ax= - (x + 2)
.r^ - 3x + 2 = 0
{x - \Y{x + 2) = 0
x=l or X = —2
The corresponding y-values are >> = — 3 and y = 0.
Points of intersection: (1, -3), (-2, 0)
74.
y = x^ - 2x2 + 1
y = 1 - x2
1 - x2 = x^ - 2x2 + 1
0 = x^ - x2
0 = x2(x + l)(x - 1)
X = - 1, 0, 1
(-1,0), (0,1), (1,0)
"^A
(0.1)/
(-1.0)/"
Yi.o)
76. y = fcc + 5 matches (b).
Use (1, 7): 7 = k{\) + 5 => fc = 2, thus, y = 2x + 5.
y = x^ + A; matches (d).
Use(l, -9): -9 = (1)^ ^k=^k= -10, thus, ); = x- - 10.
y = fcc^''2 matches (a).
Use (1, 3): 3 = kiXf''^ =^ k = 3, thus.y = 3x^/2.
xy = fc matches (c). ' •
Use (1, 36): (1)(36) = it => A: = 36, thus, xy = 36.
78. (a) Using a graphing utility, you obtain
y = -0.1283t2 + 11.0988f + 207.1116
(c) For the year 2004, r = 54 and
y = 432.3 acres per farm.
(b) 500
80. (a) If (x, y) is on the graph, then so is (— x, y) by y-axis symmetry. Since (— x, y) is on the graph, then so is (— x, —y) by
X-axis symmetry. Hence, the graph is symmetric with respect to the origin. The converse is not true. For example, y = x^
has origin symmetry but is not symmetric with respect to either the x-axis or the y-axis.
(b) Assume that the graph has x-axis and origin symmetry. If (x, y) is on the graph, so is (x, — y) by x-axis symmetry. Since
(x, -y) is on the graph, then so is (-x, - (-y)) = (-x, y) by origin symmetry. Therefore, the graph is symmetric with
respect to the y-axis. The argument is similar for y-axis and origin symmetry.
82. True
84. True; the x-intercept is
Section P.2 Linear Models and Rates of Change 287
Section P.2 Linear Models and Rates of Cliange
2. m = 2
4. OT = - 1
£ 40
6. m = -T
10. m
4-2
-2 - 1
_2
3
J.-2. 4)
-2 -(-2)
12. m = ; = 0
4—1
H 1 1 1 l-»-i
(3,-2) (4,-2i
14. m
(3/4) -(-1/4)
(7/8) - (5/4)
1 8
-3/8
16. Since the slope is undefined, the line is vertical and its equation is x = —3. Therefore, three additional points are (—3. 2),
(-3, 3), and (-3, 5).
18. The equation of this hne is
y + 2 = 2(.v + 2)
y = 2j: + 2,
Therefore, three additional points are (-3, -4), (- 1, 0), and (0, 2).
20. (a) Slope =
^ = 1
Ax ~ 3
By the Pythagorean Theorem.
.r= = 30^ + 10^ = 1000
.t = 31.623 feet.
22. (a) m = 400 indicates that the revenues increase by 400 in one day.
(b) m = 100 indicates that the revenues increase by 100 in one day.
(c) m = 0 indicates that the revenues do not change from one day to the next.
288 Chapter P Preparation for Calculus
24. 6;c - Sy = 15
6 -,
Therefore, the slope is m = j and the y-intercept is
(0, -3).
26. y = - 1
The line is horizontal. Therefore, the slope is wz = 0 and
the y-intercept is (0, - 1).
28. jc = - 1
x+ 1 =0
(-1,2)
30.
y = 4
4 = 0
-3 -2 -1
(0,4)
32. y - 4 = -f (x + 2)
5y - 20 = -Ix - 6
3;c + 5y - 14 = 0
4 - (-4) 8 -
y - 4 = 2(;c - 1)
y - 4 = Zx - 2
0 = Ix - y + 2
38. /n =
6-2
-4 -3 -2-1 12
40. wi = 0
y=-2
y + 2 = 0
-I — I — I — I-
12 3 4
(1, -2) (3. -2)
42. m =
(3/4) -(-1/4)
(7/8) - (5/4)
1 8
-3/8
1
^ + 4 3V- 4/
12y + 3 = -32jc + 40
32;c + 12y - 37 = 0
44. m — —-
y = — X + b
a
~x + y = b
a b
Section P.2 Linear Models and Rates of Change 289
= I
— Zx y
~1 2
lx + y= -1
3x + y + 2 = 0
. ^ + ^ =
1
a a
-3+1 =
1
a a
]_ _
1
a
a =
1 => .r + V = 1
X + :y - 1 = 0
50. X = 4
;c-4 = 0
52. y = \x-\
3>' - ;c + 3 = 0
54. y - 1 = 3(;c + 4)
>> = 3;c + 13
56. .t + 2v + 6 = 0
y ~ ~2^ " 3
58.
The lines do not appear perpendicular.
y
/
/\
The lines appear peqjendicular.
The lines are perpendicular because their slopes 2 and -5 are negative reciprocals of each other.
You must use a square setting in order for perpendicular lines to appear perpendicular.
290 Chapter P Preparation for Calculus
60. X + V = 7
y = -x + 7
m= -\
(a) y-2=-lU + 3)
y-l= -x-l
X +y + \ =0
(b) y - 2 = lU + 3)
y - 2 = ;c + 3
x-y + 5 = 0
62.
Ix + Ay
= 7
y
= -!.+
7
4
m
3
~ 4
(a)
y-4 =
-|(x
+ 6)
4y - 16 =
-3;c
- 18
3x +
4^ + 2 =
0
(b)
y-A
= !u
+ 6)
3y- 12
= 4jc + 24
4x-
3y + 36
= 0
64. (a) y = 0
(b);c=-l=>.:c+l=0
66. The slope is 4.50.
Hence, V = 4.5(? - 1) + 156
= 4.5r + 151.5
68. The slope is -5600. Hence, V = -5600(f - 1) + 245,000
= -5600f + 250,600
70.
^
fV'3-0'
l\
1\
You can use the graphing utility to determine that the points of intersection are (0, 3) and (3, 0). Analytically,
X- - Ax + 3 = -x^ + 2x + 3
2jc' — 6jr = 0
2x(x - 3) = 0
X = 0 ^ y = 3 => (0, 3)
• ;c = 3 ^ y = 0 =» (3, 0).
The slope of the line joining (0, 3) and (3, 0) is m = (0 - 3)/(3 - 0) = — 1. Hence, an equation of the line is
>-- 3 = -1(a:- 0)
y = -x + 3.
72.
/Kl
-6
7 -
- 4
0
10
7
mj
11 -
-5
- 4
- 0
7
5
m,
# /Wj
The points
are
not coUinear.
Section P.2 Linear Models and Rates of Change 291
74. Equations of medians:
y = -x
3a + b
c
(x + a)
(x-a)
■' -3a + b
Solving simultaneously, the point of intersection is
m-
(-a.O) , (0.0) (a.O)
(b c\ I a^ - lP-\
76. The slope of the line segment fi-om I -, - 1 to I fc, I is:
[(g^ - b'-)/c\ - (c/3) (3a" - 3fc^ - c^)/(3c) 3a' ~ 3b^ - cr
b - (b/3)
{2b)/3
2bc
The slope of the line segment from I -, - I to ( 0,
b c
2c
[(-g" + fc- + c^)/(2c)] - (c/3) _ (-3a" + 31^ + 3c- - lc')/{bc) 3a' - 3b' - c'
"^ 0-(V3)
Therefore, the points are collinear.
-bl3
2bc
78. C = 0.34x + 150. If .r = 137, C = 0.34(137) + 150 = $196.58
80. (a) Depreciation per year:
^ = $175
>- = 875 - 175a:
where 0 < x < 5.
db) y = 875 - 175(2) = $525
82. (a) V = 18.91 + 3.97.i: (x = quiz score, >> = test score)
(b) "M
(c) 200 = 875 - 175x
175;c = 675
X = 3.86 years
(c) ]fx = 17,y = 18.91 + 3.97(17) = 86.4.
(d) The slope shows the average increase in exam score
for each unit increase in quiz score.
(e) The points would shift vertically upward 4 units. The
new regression line would have a y-intercept 4 greater
than before: v = 22.91 + 3.97.r.
84. 4.1 + 3y - 10 = 0 => c/
|4(2) + 3(3) - 10| _ 7
Va^TY- ~ 5
M. x+ \=Q^>d =
|1(6) + (0)(2) + 1|
^1- + 0"
= 7
88. A point on the line 3x - 4.v = 1 is (- 1, - 1). The distance from the point (- 1, - 1) to 3.v - 4v - 10 = 0 is
1-3 + 4- 10| 9
292 Chapter P Preparation for Calculus
90. y = ntx + 4=>mx + (-\)y + 4 = 0
|Aci + gy, + C\ ^ \m3 + (- 1)(1) + 4|
^ |3m + 3|
Vm^ + 1
The distance is 0 wlien w = — 1. In this case, the line y = —x + 4 contains the point (3, 1).
92. For simplicity, let the vertices of the quadrilateral be
(0, 0), (a, 0), (b, c), and (d, e), as shown in the figure. The
midpoints of the sides are
l»
a + bc\lb + dc + e\ ^Id e
The slope of the opposite sides are equal:
c + e
0
a + b
b + d
£
-1 = £
d b
0
c + e
a + b b + d
111 1
Therefore, the figure is a paralleogram.
(^■^)
94. If ffZi = — l/m,, then mjOTj = ~ 1- Let Lj be a line with
slope mj that is perpendicular to Ly Then miWij = — 1.
Hence, m2 = m-^=^ L^ and Lj are parallel. Therefore, Lj
and Z,j are also perpendicular.
96. False; if m, is positive, then m^= - l/m^ is negative.
Section P.3 Functions and Their Graphs
2. (a) f(-l) = V-2 + 3 = yr = 1
(b) /(6) = V6T3 = 79 = 3
(c) /(c) = v^n
(d) /(jc + Ax) = Jx + Kx + 3
45
4. (a) ^(4) = 4^(4 - 4) = 0
(b) ^(1) = (im - 4) = !(-!)
(c) gic) = c^{c - A) = c^ - Ac^
(d) g(t + 4) = (r + 4)2(f + 4-4)
= (r + A)h = fi + %t^+\6t
6. (a) /(tt) = siniT = 0
(lTr\ . (Itt
(Of
ns
73
2
(b)/l
t) = ^^l'
4
■v^
8.
/(;c) -/(I) 3;c - 1 - (3 - 1) _ 3(;c - 1)
1
1
1
= 3,x^ 1
10. — — = ; = ; = x[x + \),x i= 1
X — \ X — I X — 1
Section P. 3 Functions and Their Graphs 293
12. gix) =x^-5
Domain: (— oo, oo)
Range: [-5, oo)
14. h{t) = cot t
Domain: all t i^ k-rr, k an integer
Range: (-00,00)
16. g{.
X - 1
Domain: (-00, 1), (1, 00)
Range: (-00, 0), (0, 00)
18. f(x) =
x'^ + 2,x < \
2jf2 + 2, ;c> 1
(a) /(-2) = (-2)= + 2 = 6
(b) /(O) = 02 + 2 = 2
(c) /(I) = 1^ + 2 = 3
(d) f{s^ + 2) = 2(52 + 2)2 = 2j^ + 8^2 + 10
(Note: j2 + 2 > 1 for all s)
Domain: (—00, 00)
Range: [2, 00)
20. fix) =
Vx + 4,x < 5
(x - 5)2, X > 5
(a) /(-3) = V-3 + 4 = yr = 1
(b) /(O) = VoT^ = 2
(c) /(5) = ym = 3
(d) /(lO) = (10 - 5)2 = 25
Domain: [-4, 00)
Range: [0, 00)
22. g(x) = -
Domain: (-00, 0). (0, 00)
Range: (- 00, 0), (0, 00)
24. fix) = ;x3 + 2
Domain: (—00, 00)
Range: (—00, 00)
26. fix) =x+ y4-x2
Domain: [—2, 2]
Range:
[-2,272] ==[-2,2.83]
y-intercept: (0, 2)
x-intercept: ( - V2, o)
(-v/2.0>
28. Me) = -5cos^
Domain: (-00, 00)
Range: [-5,5]
30. 7x2-4
0:
V^
y is a function of x. Vertical lines intersect the graph
at most once.
32. x2 + v2 = 4
y = ±74 - X-
y is not a function of x. Some vertical lines intersect
the graph twice.
34. x2 + y = 4 => y = 4 — .r2
y is a function of x since there is one value of y for
each X.
36. xh- - .r2 + 4v = 0 =
.r2 + 4
y is a function of x since there is one value of y for
eachx
294 Chapter P Preparation for Calculus
38. Piix) = x' - X + 1 has one zero. /JjW = x^ - x has
three zeros. Every cubic polynomial has at least one zero.
Given p(x) = Ax^ + Sx^ + Cx + D, we have p -> - oo as
x-^ — oo andp— >oo asx— >oo if i4 > 0. Furthermore,
p— >oo asx— > — oo and/3— > — oo as X— >oo if A < 0.
Since the graph has no breaks, the graph must cross the
X-axis at least one time.
r^
40. The function is/(x) = cx. Since (1, 1/4) satisfies the
equation, c = 1/4. Thus,/(x) = (l/4)x.
42. The function is h(x) = c-y\x\. Since (1, 3) satisfies the
equation, c = 3. Thus, h(x) = 3 V]x[.
2 — 0 1
44. The student travels _ = - mi/min during the first 46. (a)
4 minutes. The student is stationary for the following
6-2
2 minutes. Finally, the student travels
during the final 4 minutes.
10
1 mi/min
,.
500-
^
400-
X
^
V
300-
y
200'
f'
100-
10
20
30
40
50
(b) A(15) == 345 acres/farm
48. (a) g(x)=/(x-4)
g(6)=/(2) = 1
g(0)=/{-4) = -3
Shift /right 4 units
(b) g(x)=/(x + 2)
Shift/left 2 units
(c) g(x)=/(x)+4
Vertical shift upwards
4 units
(d) g(x)=/(x)- 1
Vertical shift down 1 unit
(e) g{x) = 2/(x)
g{2) = 2/(2) = 2
g(-4) = 2/(-4) =
(0 gW = 5/W
g(2) = ^/(2) = i
g(-4) = i/{-4) =
(-4.-6)
50. (a) h(x) = sin(x -I- (n/l)) + 1 is a horizontal shift ir/2 units to the left, followed by a vertical shift 1 unit upwards.
fb) h(x) = - sin(x - 1) is a horizontal shift 1 imit to the right followed by a reflection about the x-axis.
Section P. 3 Functions and Their Graphs 295
52. (a)/(^(l))=/(0) = 0
(b) g(f(\)) = g(l) = 0
(c) g{fm = g{Q) = - 1
(d)/(5(-4))=/(15)= 715
(e)/(^W)=/(x2- \) = J^^X
(f) gifbc)) = g{J~x) = {J~xf - \ = .X - \ {x > 0)
54. fix) = X' - ).,g(x) = cosj:
{f'g){x)=Mx)) =/(cos.v) = cos^v - 1
Domain: (—00,00)
{g'f){x)=g{x^- l) = cos(x2- 1)
Domain: ( — 00, ca)
NoJ'g^ g'f.
56. (/og)(A) =f{V^T2) = -^1=
~Jx + 2
Domain: (—2, 00)
=/)W
.^=7^
2x
You can find the domain of g °/by determining the intervals where (1 + 2x) and x are both positive, or both negative.
-2 -1 _i 0 1 2
Domain: ( — 00, — jj, (0, 00)
58. (a) 25
60. f(-x) = If^ = - l/^ = -fix)
Odd
(b) H'
1.6
0.0021 ^)- + 0.005(^
0.029
= 0.000781 25a- 2 + 0.003125a - 0.029
62. /(-a) = sin^(-.v) = sin(-A) sin(-.t) = (-sin.r)(-sin a) = sin^A
Even > (
64. (a) If/ is even, then (-4, 9) is on the graph.
66. /(-a) = a,„(-Af " + a2„_2(-AP"-^ + • ■ ■ + ^.(-a-)^ + a^
(b) If/ is odd, then (-4, -9) is on the graph.
= /(-v)
Even
68. Let F(a) = /(.v)g(.v) where/is even and g is odd. Then
F(-a) =/(-.v)g(-A) =/(A)[-g(A)] = ~fix)g(x) = -Fix).
Thus, Fix) is odd.
296 Chapter P Preparation for Calculus
70. (a) Let F{x) = fix) ± g{x) where/and g are even. Then, F{-x) = f(~x) ± g{-x) = f{x) ± g(x) = F(x).
Thus, F{x) is even.
(b) LetFW =fix) ± ^W where fandg are odd. Then, F(-x) = f{- x) ± g(- x) = -f(x) + g{x) = -F{x).
Thus, F{x) is odd.
(c) Let FW = fix) ± gix) where f is odd and g is even. Then, F(-x) = fi-x) ± gi-x) = -fix) ± gix).
Thus, Fix) is neither odd nor even.
72. By equating slopes.
y-2 _ 0- 2
0-3 X- 3
6
X- 3
6
74. True
y-2
y =
x-2,
+ 2 =
2x
X- ^'
= Jx^ + y^ = -v/jc^ +
Ix \2
X-2,
76. False; let/W = x^. Then/(3x) = ilxf = 9x^ and 3/W = 3;^^. Thus, 3/U) ^ /(3x).
Section P.4 Fitting Models to Data
2. Trigonometric function
4. No relationship
6. (a) 20
No, the relationship does not appear to be linear.
(b) Quiz scores are dependent on several variables such as
study time, class attendance, etc. These variables may
change from one quiz to the next.
8. (a) s = 9.1t + 0.4
(b) 45
The model fits well,
(c) If / = 2.5, s = 24.65 meters/second.
10. (a) Linear model: H = -0.3323/ + 612.9333
(b) 600
The fit is very good,
(cj When t = 500,
H = -0.3323(500) + 612.9333 « 446.78.
12. (a) S = 180.89jc2 - 205.79;c + 272
(b) 25000
(c) When x = 2, 5 = 583.98 pounds.
Review Exercises for Chapter P 297
14. (a) t = 0.0027 1^2 - 0.05295 + 2.671
(b) 21
(c) The curve levels off for s < 20.
(d) t = 0.002^2 + 0.0346i + 0.183
The model is better for low speeds.
18. (a) H{t) = 84.4 + 4.28 sinf ^ + 3.86
One model is
TTt
C{t) = 58 + 27sin(— + 4.1
(b) 100
20. Answers will vary.
16. (a) T= 2.9856 x lO'V^ _ 0.0641 /j^ + 5.2826p + 143.1
(b) 350
(c) For T = 300°F, p = 68.29 pounds per square inch.
(d) The model is based on data up to 100 pounds per
square inch.
(C) 100
(d) The average in Honolulu is 84.4.
The average in Chicago is 58.
(e) The period is 12 months (1 year).
(f) Chicago has greater variability (27 > 4.28).
Review Exercises for Chapter P
2. y = (.t - l)(.t - 3)
;r = 0 ^ .V = (0 - 1)(0 - 3) = 3 ^ (0, 3) v-intercept
;y = 0 ^ 0 = U - l)(x - 3) => .V = 1, 3 => (1, 0), (3, 0) A-intercepts
4. XV = 4
X = 0 and y = 0 are both impossible. No intercepts.
6. Symmetric with respect to -v'-axis since
y = (-x)-* - (-x)- + 3
v = x" - x~ + 3.
298 Chapter P Preparation for Calculus
8. 4x -2y = 6
y = 2x - 3
Slope: 2
y-intercept: —3
10. 0.02X + 0.15y = 0.25
2x + 15y = 25
12. y = x{6- x)
y = -■i5-« +
Slope: -fj
y-intercept: j
14. y = |;c - 4| - 4
16. y = Si/x
Xmin = -40
Xmax = 40
Xscl = 10
Ymin = -40
Ymax = 40
Yscl = 10
18. • y = ;t -H
(x+ I) -x^ = 7
0 = x'- - a: 4- 6
No real solution
No points of intersection
The graphs of y = Jt -I- 1 and
y = x^ + 7 do not intersect.
20. y = kx^
(a) 4 = kilf =^k = 4andy = 4x^
(c) 0 = M0)3 => any A- will do!
22.
12-
10-
8--
S-
4-
2--
H — I — I — I — I h-
12 3 4 5 6
(7, 12)
—h-»-'
(7,-1)
The line is vertical and has no slope.
24.
(b) 1 = k(-2y =^k= -^ andy = -jx^
(d) -I = k{-iy=^k= l=^y = x^
3-(-l) 3-6
-3 - r -3 -
4 -3
-3-f -11
-44 = 9 -^ 3r
-53 = 3r
Review Exercises for Chapter P 299
26. J, - 6 = OU - (-2))
y = 6 Horizontal line
(-2,6)
I I I I I I > -
28. m is undefined. Line is vertical.
x = 5
6--
4
-4-2 - ■ 2 4
-2--
(5.4)
30. (a)
y-3
r(-t - 1)
Sy - 9 = -2;c + 2
2x -H 3y - 1 1 = 0
(b) Slope of perpendicular line is 1.
>- - 3 = lU - 1)
y = X + 2
Q= X- y + 2
4-3 .
(c)
2 - 1
y - 3 = l(x - 1)
y = x + 2
Q = X- y + 2
(d) y = 3
y- 3 = 0
32. (a) C = 9.25r + 13.50f + 36,500
= 22.75f + 36,500
(b) /? = 30f
(c) 30r = 22.75r + 36,500
7.25r = 36,000
r « 5034.48 hours to break even.
34. X- - V = 0
Function of x since there is one value for v for each x.
36. X = 9 - r
Not a function of x since there are two values of y for
some x.
38. (a) /(-4) = (-4)2 + 2 = 18 (because -4 < 0)
(b) /(O) = |0 - 2| = 2
(c)/(l)= |1 -2| = 1
40. f(x) = 1 - .r= and g{x) = Iv + 1
(a) f[x) - gix) = (1 - .r^) - (2r + \) = -x~ - Ix
(b) f(x)g{x) = (1 - .v-)(2v + 1) = -It^ - .r= + Ir -I- 1
(c) g{f{x)) = ^(1 - .X-) = 2(1 - .v^) + 1 = 3 - li-
300 Chapter P Preparation for Calculus
42. f(x) = x^ -3.xP-
(a) The graph of g is obtained from/by a vertical shift
down 1 unit, followed by a reflection in the jc-axis:
gix) = -[/W - 1]
= -x^ + 3x^+ I
(b) The graph of g is obtained from /by a vertical shift
upwards of 1 and a horizontal shift of 2 to the right.
g(x)=f{x-2) + 1
= (x- 2)3 - 3(x -2)2+1
44, (a) fix) = x^x - 6f
(b) gix) = Jt3(x - 6)2
300
Ai
^ /
(c) hix) = x^U - 6)3
200
4 ■ — ^-r— . ,./ , . 10
46. For company (a) the profit rose rapidly for the first year,
and then leveled off. For the second company (b), the
profit dropped, and then rose again later.
Problem Solving for Chapter P
48. (a) y = - 1.204.t + 64.2667
(b) 70
(c) The data point (27, 44) is probably an error.
Without this point, the new model is
y = -1.4344JC + 66.4387.
2. Let y = mx + 1 be a tangent line to the circle from the point (0, 1). Then
x^ + iy+l)^=l
x^ + imx+ I + l)^= \
irrfi + \)x'^ + 4mx + 3 = 0
Setting the discriminant b^ — 4ac equal to zero,
16m^ - Airrr + 1)(3) =0
\6m- - I2m^ = 12
4m^ = 12
m = ±y3
Tangent lines: y = V^x + 1 and v = - V^x + 1.
Problem Solving for Chapter P 301
4. (a) /(x + 1)
(b) f(x) + 1
(c) lf{x)
(d) f{-x)
(e) -/W
(0 i/wi
4-
:Y^^y:
(g) /(ki)
6. (a) 4y + 3;c = 300 => v
300 - 3a-
w ^ /-, . /'300 - 3.x\ -3.V- + 300x
Aix) = x(2y) = xl^ j =
Domain: 0 < x < 100
(b)
25 50 75 100
(c) A(x) = --{x- - IOOa)
= --(x^ - lOOx + 2500) + 3750
Maximum of 3750 ft" at x = 50 ft, >- = 37.5 ft.
= --(x - 50)= + 3750
A(50) = 3750 square feet is the maximum area,
where x = 50 ft and v = 37.5 ft.
302 Chapter P Preparation for Calculus
8. Let d be the distance from the starting point to the beach,
distance
Average velocity =
time
2d
120 60
120 60
= 80 km/hr
12 3 4 5
3-21 1
(a) Slope = = - Slope of tangent line is greater than —
2-11 1
(b) Slope = _ = - Slope of tangent line is less than -.
, , ^, 2.1 - 2 10 ^, ^ ,. . ^ 10
(c) Slope = — = — . Slope or tangent Ime is greater than — .
(d) Slope = (4 + ;,)_4
vm- 2
, , V4 + /! - 2 V4 + /! - 2 V4 + /! + 2
(e) = • — ==
h h V4 + /! + 2
_ (4 + /;) - 4
//(V4 + /I + 2)
1
JA + h + 2
hi-Q
As /! gets closer to 0, the slope gets closer to — . The slope is - at the point (4, 2).
Problem Solving for Chapter P 303
12. (a)
kl
U-4)2+/ = feV + )^)
(A:2 - ly + {k^ - l)f + &x = 16
If k = 1, then ;c = 2 is a vertical line. So, assume /c' - 1 ¥= 0. Then
( 4 V , 16 16
U + t; — 7 1 +y^ = - r +
k^- 1
x +
A:2- 1
+ / =
fc2 - 1 (F - 1)2
Ak
k^- 1
, Circle
(b) Ift = 3, (;t + ^)" + / = (!)'
(c) For large k, the center of the circle is near (0, 0), and the radius becomes smaller.
14. fix) =y =
1
1 - X
(a) Domain: all -t t^ 1
Range: all v t^^ 0
1
(b) fifix)) =/l
1 - x
1 \ - X x-\
1 — .V — 1 —X X
\ - x) \ - X
Domain: all x i^ 0, 1
(x- 1
(c) f{f{f{x)))=f\
1
1
X j (x- W 1
X I X
Domain: all x t^ 0, 1
(d) The graph is not a line. It has holes at (0, 0) and (1, 1).
{ 1 — »-^
CHAPTER 1
Limits and Their Properties
Section 1.1 A Preview of Calculus 305
Section 1.2 Finding Limits Graphically and Numerically 305
Section 1.3 Evaluating Limits Analytically 309
Section 1.4 Continuity and One-Sided Limits 315
Section 1.5 Infinite Limits 320
Review Exercises 324
Problem Solving 327
CHAPTER 1
Limits and Their Properties
Section 1.1 A Preview of Calculus
Solutions to Even-Numbered Exercises
2. Calculus: velocity is not constant
Distance = (20 ft/sec)(15 seconds) = 300 feet
4. Precalculus: rate of change = slope = 0.08
6. Precalculus; Area = tt[J2)'
8. Precalculus: Volume = 7r(3)-6 = 54tt
10. (a) Area "5+- + - + -« 10.417
(b) You could improve the approximation by using more rectangles.
Section 1.2 Finding Limits Graphically and Numerically
X
1.9
1.99
1.999
2.001
2.01
2.1
fix)
0.2564
0.2506
0.2501
0.2499
0.2494
0.2439
lim ^ 7 = 0.25 (Actual limit is 3.)
x-^2 X 4
4.
X
-3.1
-3.01
-3.001
-2.999
-2.99
-2.9
fix)
-0.2485
-0.2498
-0.2500
-0.2500
-0.2502
-0.2516
lim :; «= -0.25 (Actual limit is -j)
->-3 .T + 3 ^
6.
X
3.9
3.99
3.999
4.001
4.01
4.1
fix)
0.0408
0.0401
0.0400
0.0400
0.0399
0.0392
lim ^^ —, =» 0.04 (Actual limit is ^.)
j:->4 X - 4
X
-0.1
-0.01
-0.001
0.001
0.01
0.1
fix)
0.0500
0.0050
0.0005
-0.0005
-0.0050
-0.0500
cos X — I
lim '■ « 0.0000 (Actual limit is 0.) (Make sure vou use radian mode.)
.C-.0 X
305
306 Chapter 1 Limits and Their Properties
10. lim (;c^ + 2) = 3
jr-»l
12. lim/(x) = lim (;c^ + 2) = 3
14. lim
1
■ does not exist since the
3 a: - 3
function increases and decreases
without bound as x approaches 3.
16. lim sec X = 1
18. lim sin(7rx) = 0
20. C{t) = 0.35 - 0.121- (f- 1)
(a)
(b)
I
3
3.3
3.4
3.5
3.6
3.7
4
C{t)
0.59
0.71
0.71
0.71
0.71
0.71
0.71
lim CW = 0.71
(c)
t
3
2.5
2.9
3
3.1
3.5
4
C{t)
0.47
0.59
0.59
0.59
0.71
0.71
0.71
lim C{t) does not exist. The values of C jump from 0.59 to 0.71 at f = 3.
f— *3.5
22. You need to find S such that 0 < |x — 2| < 5 implies
[fix) - 3| = 1^2 - 1 - 3| = 1^2 - 4| < 0.2. That is,
-0.2 < x2 - 4 < 0.2
4 - 0.2 < x^ < 4 + 0.2
3.8 < x^ < 4.2
JJI < X < JA2
Vl8 - 2 < X - 2 < V4I - 2
So take S = 742 - 2 « 0.0494.
ThenO < |x - 2| < 8 implies
-{742 - 2) < X - 2 < 742 - 2
73^ - 2 < X - 2 < 742 - 2.
Using the first series of equivalent inequalities, you obtain
|/(x) - 3| = |x2 - 4| < 0.2.
2^- IT. '-2
-f
-f
f (. - 4)
< 0.01
< 0.01
< 0.01
0 < |x - 4| < 0.02 = 5
Hence, if 0 < |x - 4| < 8 = 0.02, you have
\U - 4)
-f
< 0.01
< 0.01
< 0.01
|/(x) -L\< 0.01
Section 1.2 Finding Limits Graphically and Numerically 307
26. lim (x^ + 4) = 29
\(x^ + 4) - 29| < 0.01
|a:2 - 25| < 0.01
|(x + 5){x - 5)1 < 0.01
0.01
5 <
U + 5
If we assume 4 < x < 6, then 5 = 0.01/11 = 0.0009.
Hence, if0<|jr-5|<6 = —rr-, you have
1
|.x- 5||a; + 5| < 0.01
|;c2 - 25| < 0.01
(x^ + 4) - 29| < 0.01
|/(x) - L\ < 0.01
(0.01)
28. lim (2;c + 5) = - 1
a:— >-3
Given e > 0:
|(2r + 5)-(-l)| < e
|2x + 6| < e
2\x + 3| < 6
|;t + 3| <|=6
Hence, let 5 = e/2.
Hence, if 0 < |jr + 3| < 5 = -, you have
1^ + 31 < I
i2x + 6| < e
|(2x + 5)-(-l)| < 6
l/W -L\<e
30. lim %x + 9) = 5(1) + 9 = f
Given e > 0:
|V3^ + 9j - y| < 6
i|x-l|<e
k-l|<56
Hence, let 6 = (3/2)e.
Hence, if 0 < |j: — 1 1 < S = je, you have
\x - \\ < ^€
32. lim(-l) = -1
JC— ^2
Given e > 0: |-1 - (-1)| < e
0 < e
Hence, any 5 > 0 will work.
Hence, for any 5 > 0, you have
|{-l)-(-l)| <e
l/W - i| < e
|(|j: + 9)-f| < e
l/W - L| < 6
34. lim v^ = 74 = 2
Jr— >4
Given e > 0: |V^ - 2| < e
|V5 - 2| |v^ -I- 2I < e| V^ + 2I
|x - 4| < e|7c + 2I
Assuming 1 < a- < 9, you can choose 5 = 3e. Then,
0 < \x- A\ < 5 = 3e => |.r - 4| < e| Va + 2I
=> ITx- - 2I < e.
36. lim |.v- 3| =0
Given 6 > 0:
|(.r - 3) - 0| < e
|.T- 3| < e= 5
Hence, let 8 = e.
Hence for 0 < |.r - 3 1 < 6 = e. you have
|.v - 3| < 6
||..-3| -0| <6
|/(.t) -L\<e
308 Chapter 1 Limits and Their Properties
38.
lim {x^
Given e
+ 3x)
> 0:
= 0
\{^
+ 3x) -
- 0|
<
e
\x{x^
3)1
<
6
k-
^-3|
<
w
If we assume - 4 < x < — 2, then 5 = e/4.
Hence for 0 < |j: — (— 3)| < 5 = t. you have
\x + 3\ <-e < i-j-e
' A \x\
\x{x + 3)1 < e
|x2 + 3;t - 0| < 6
l/W - L| < e
40. /W =
x-Z
x^ - Ax + l,
V
h
lim/W = -
The domain is all x i= 1, 3. The graphing utility does not
show the hole at (3, 5).
42. fix) = V4
x'^ — 9
Mm fix) = -
j:— »3 0
3
9 '
44. (a) No. The fact that/(2) = 4 has no bearing on the exis-
tence of the limit of /W as x approaches 2.
(b) No. The fact that lim/(x) = 4 has no bearing on the
value of /at 2.
The domain is all a: # ± 3. The graphing utility does not
show the hole at (3, g).
46. Let p(x) be the atmospheric pressure in a plane at
altitude x (in feet).
lim p(x) = 14.7 lb/in2
48. 0.002
Using the zoom and trace feature, 5 = 0.001. That is, for
0 < Ix - 21 < 0.001,
x-2
- 4
< 0.001.
50. True
52. False; let
f{x) =
Ax, x + A
10,
lim/(x) = lim (x2 - 4a;) = 0 and/(4) = 10 ^t 0
j:-»4 jr->4
54. lim
^
\2
J:->4 X - a
= 1
n
4 + [0.1]"
/(4 + [0.1]")
1
4.1
7.1
2
4.01
7.01
3
4.001
7.001
4
4.0001
7.0001
n
4 - [0.1]"
/(4 - [0.1]")
1
3.9
6.9
2
3.99
6.99
3
3.999
6.999
4
3.9999
6.9999
Section 1.3 Evaluating Limits Analytically 309
56. fix) = mx + b, m i^ Q.l^X e > Qhe given. Take 5 = -j — r.
\m\
IfO <\x- c\ < 8 = 7—7, then
\m.\\x — c\ < e
\mx — mc\ < e
\{mx + b) - (mc + b)\ < e
which shows that lim (mx + b) = mc + b.
58. lim g{x) = L,L > 0. Let e = jL. There exists 5 > 0
such that 0 < |a: - 0| < 8 implies \g{x) - L\ < e = jL.
That is,
-5L < gix) - L < {l
\l < gix) < jL
Hence for x in the interval ic — S, c + S), x ¥= c,
gix) >\l>Q.
Section 1.3 Evaluating Limits Analytically
2. 10
(a) lim gix) = 2.4
(b) lim^U) = 4
x—*0
(a) lim/(f) = 0
/->4
(b) lim /(f) = -5
«W = \._a
.t-9
/(r) = fk - 4|
6. lim x3 = (-2)3 =
a:-»-2
8. lim (3x + 2) = 3(-3) + 2 = -7
x-»-3
10. lim(-r^+ 1) = -(1)^+ 1 = 0
X-*l
2 2
14. lim
3X + 2 -3 + 2
-2
12. lim (3.^3 - 2r= + 4) = 3(1)^ - 2(1)" + 4 = 5
:t-^3 ;c + 5 3 + 5 8
ifi I- V^^^ V3 + 1 ,
18. Imi — = -z :- = -2
a:->3 X - 4 3-4
20. lim4/;c + 4 = ^4 + 4 = 2
x->4
22. lim i2x - ly = [2(0) - 1]' = -1
24. (a) lim fix) = (-3) + 7 = 4
(b) lim gix) = 4^ = 16
(c) lim gifix)) = ^(4) = 16
x—*-3
26. (a) lim/(jc) = 2(4^) - 3(4) + 1 = 21
(b) lim gix) = ^21 +6 = 3
(c) \\mgifix)) = gi2l) = 3
x-*4
28. lim tan X = tan 77 = 0
x-*ir
30. lim sm ^:— = sm — = 1
j:-»1 2 2
32. lim cos 3x = cos Stt = - 1
x—*lr
34. lim cos X = cos ^;r = X
j->57r/3 3 2
« ,. /tT.xA 777 -2v^
36. lim sec —- = sec — - = — :; —
j-»7 V 6 / 6 3
310 Chapter 1 Limits and Their Properties
38. (a) lim [4/W] = 41im/(;c) = 4 - = 6
x^c j:— *c \Z/
(b) lim [fix) + g{x)] = lim/(x) + lim g{x) = ^ + \=2
x—^c j:-+c x-*c L L
(c) lim \S{x)g{x)-\ = riim/Wiriim g{x)\
IB=!
40. (a) XxmlfJQ) = 3/lim/(x) = l/in = 3
Jf-»C
(c) lim [/W]2 = [lim/(x)l2 = (27)2 ^ 729
(d) lim[/(;c)]2/3 ^ riin,/(;c)]2/3 = (27)2/3 = 9
X^ - Zx
42. /W = j: — 3 and h{x) = agree except at j: = 0
1 ;c
44. g{x) = 7 and/U) = -; agree except at j; = 0.
X — I x'- — X
(a) lim h{x) = lim f{x) = -5
jr->-2 jc-*-2
(b) lim/iW = lim/W = -3
(a) lim/(x) does not exist.
jr-»l
(b) lim/(;c)= -1
JC-^O
2;t2 - X - 3
46. /(x) = — and g{x) = 2x — 3 agree except at
JC^ + 1
48. /(j:) = — — — and g{x) = j? — x -\- 1 agree except at
x= -1.
lim f{x) = lim g{x) = -5
lim f{x) = lim g(x) = 3
J—* - 1 JT— » - 1
/
/[
\
,. ,. 1- X -{x- 2)
50. lim -:; = lim • ^ —
;c-*2;c2-4 x^2 (a: - 2)(a: + 2)
= lim
-1
j:->2 X -V 1
^^ ,. ;c- - 5;c + 4 ,. (x - 4)(;c - 1)
52. hm — — = hm 7 -(7 -(
x->4 x2 - 2x - 8 x^A [x - 4)(;c + 2)
= lim 7 rr = ~ = x
x-»4 U + 2) 6 2
,. ,. JlVx-Ji ,. V2 + a; - V2 V2 + ;t + v^
54. lim = lim • — , ^
^^0 X x-^ X Jl + X + J2
2+X-2 ,. 1
= lim 7 — , ;=A— = lim
1 72
^-»o(V2 + X+ J7)x ^-^0 J2 + X + J2 ijl 4
,, ,. v7TT-2 ,. Vx + 1 - 2 Vx + 1 + 2 ,.
56. hm ; = lim ;: • — , — - = hm ■
A- 3
lim -
^**" -^ ~ I'l" -^- . T^ — 11111 , TTr ; TT — 11111 , T"
x-^i X - 3 x^i X - 3 Vx+ 1 + 2 -.-^3 (x - Z\jx + 1 + 2] :c^3 vTTT + 2
_J j_ 4 - (x + 4)
.„ ,. A + 4 4 ,. 4(x + 4) ,. -1 1
58. hm = hm — ^^ '— = hm -7 — -— r = - —
x^o X x-^0 X x->o 4(x + 4) 16
„ ,. (x + Ax)2 - x2 ,. x^ + 2xAx + (Ax)2 - x2 ,. Zb;(2x + Ax) ,. ,, , . ^
60. hm -^ = hm — ^ — = hm — ^^-r = hm (2x + Ax) = 2x
Aar-^O Ax Ax->0 Ax Ax-»0 Ax Ai^O
Section 1.3 Evaluating Limits Analytically 311
^ ,. U + Ajc)3 - x3 ,. x^ + 3x'\x + 3x{Axy + (Axf - jc'
62. lim -. = lim
Ax->o Ajc '^->0 Ax
,. Ax{3x^ + 3xAx + (Ax)^) ,. ,, , , . ,. ,,, , ,
hm — ^^ ; ^^ — — = hm (3r= + 3xAx + (Ax)^) = 3x^
Aar-»0 A.V Ax->0
64. fix)
A- Jx
X - 16
X
15.9
15.99
15.999
16
16.001
16.01
16.1
/w
-.1252
-.125
-.125
7
-.125
-.125
-.1248
Analytically, lim
(4-v^
.r-16 X - 16 x^\(,[J~y_ + 4)(^ _ 4)
= lim — F = --.
-'-'IS Jx-v A 8
It appears that the limit is -0.125.
66. lim ^ = 80
j:->2 X — L
X
1.9
1.99
1.999
1.9999
2.0
2.0001
2.001
2.01
2.1
fix)
72.39
79.20
79.92
79.99
7
80.01
80.08
80.80
88.41
, , . „ ,. .r5 - 32 ,. ix - 2)(.x^ + 2x3 + 4x2 + 8x + 16)
Analytically, lim —- = lim r
■^ -^ x^2 X - 2 x^2 X - 2
= lim (x" + 2x' + 4x^ + 8x + 16) = 80.
j-»2
(Hint: Use long division to factor x^ - 32.)
3(1 - cos.x)
68. lim =^ ^^^^^ = lim [sf^-^^^)] = (3)(0) = 0
„. ,. cose tan 0 ,. sm 0
70. lim = hm — ;— = 1
8->0 0 B->0 6
_^ ,. tan"x ,. sin-x
72. lim = hm ^— = lim
x->0 X Jr->0 X cos' X .t->0
sinx sin x 1
X cos- X J
74. lim (^ sec (^ = tt(— 1) = — tt
= (1)(0) = 0
-. ,. 1 — tanx ,. cosx — sinx
76. lim -: = hm
Jr-.7r/4 sin X — cos X x-^it/4 Sin X cos X — COS" X
(sinx — cos.v)
= lim
v/4 cos x(sin X — cos x)
1
= lim
x->it/4 cosx
= lim (-secx)
.r— nr/4
= -72
_„ ,. sin 2x
78. hm . .,
x-iO sm 3x
r Vsinlv
= hm 2 -r —
-t-»o L V Iv
'-'"\]m-M>-\
312 Chapter 1 Limits and Their Properties
SO.fih) = (1 +cos2/!)
h
-0.1
-0.01
-0.001
0
0.001
0.01
0.1
m
1.98
1.9998
2
?
2
1.9998
1.98
r\/\r\
Analytically, lim (1 + cos 2h) = \ + cos(O) =1 + 1=2.
The limit appear to equal 2.
82. f(x)
X
-0.1
-0.01
-0.001
0
0.001
0.01
0.1
fix)
0.215
0.0464
0.01
?
0.01
0.0464
0.215
Analytically, lim^ = lim^f^^^) = (0)(1) = 0.
j-jO \ X
The limit appear to equal 0.
„. ,. fix + h)- fix) Jx + h- J~x .. Jx + h
84. lim ; — = lim ; = lim —
/i->o h h->o h /i->o h
lim
X + h — x
lim ■
^x + h + Jx
Jx + h + Jx
1
''-*o h[Jx + h + Jx) '"^0 Jx + h + J~x ijx
„, ,• fix + h) - fix) ,. ix + h)- - 4(x + h) - ix^ -Ax) ,. ' x^ + 2xh + h'^ - 4x - 4h - x^ + Ax
86. hm-^^ r — =^-^ = hm -^^— —, — ^^ = Iim ;
h->o h A->o h A-»o h
,. hilx + /i - 4) ,. ,^ , ,, ^
= hm -^^ ; = hm (2x + /z - 4) = 2jc - 4
88. iim [b - \x- a|] < lim/(jc) < lim [b + \x - a|]
b < Wmfix) < b
Therefore, lim/(;t) = b.
90. fix) = |.rsinji:|
6
lim brsin;c = 0
92. fix) = \x\ cos X
94. hix) = X cos —
X
:
Iim X cos X = 0
jc->0
lim X cos - = 0
x-*0 \ X
Section 1.3 Evaluating Limits Analytically 313
x^ - 1
96. fix) = — — j- and g(x) = x + \ agree at all points
except .X = 1.
98. If a function /is squeezed between two functions h and g,
h{x) < fix) < gix), and fi and g have the same limit L as
X— >c, then Vim fix) exists and equals L.
100. fix) = x, gix) = sin^ X, hix)
sm'- X
o^Tn
A
•y
When you are "close to" 0 the magnitude of g is "smaller"
than the magnitude of/ and the magnitude of g is
approaching zero "faster" than the magnitude of/.
Thus, |g|/|/| ~ Owhen jc is "close to" 0
102. sit) = - 16r2 + 1000 = 0 when r = J^ ^^^
16
seconds
.. ■'i 2 j '^^ ,. 0-(-16r2+ 1000)
lim , -p= = lim , -/==
t^sVToh 5V10 _ J. (-»5yio/2 5V10 _
16 f-
lim
125
16
lim
t + ^Y. - ^)
2 /
■»5yio/2 SyiO _ r->5yio/2
5V10
= lim
r^5yio/2
■ leff + ^^^j = -SOyiO ft/sec = -253 ft/sec
104. -4.9f- + 150 = 0 when t =
The velocity at time t = a\s
/l50 ^ /l500
4.9 ~ V 49
= 5.53 seconds.
,. sia) - sit) ,. (-4.9a^ + 150) - (-4.9r + 150) ,. -4.9(a - f)(a + r)
lim = lim = lim-
I-»n a — t r-»n a — t r->n
= lim -4.9(a + t) = -2a(4.9) = -9.8a m/sec.
a - t
Hence, if a = V 1500/49, the velocity is -9.8V1500/49 = -54.2 m/sec.
106. Suppose, on the contrary, that lim gix) exists. Then, since Vim fix) exists, so would lim [fix) + gix)], which is a
.r— *c .x—*c x—*c
contradiction. Hence, lim ^(jc) does not exist.
x-*c
108. Given /(.x) = x", n is a positive integer, then
limx" = limCxx""') = [limJflimx""']
= c[lim ixx"-^ = c[lim.x][limx"--]
= c(c)lim (xa:""') = • ■ ■ = c".
110. Given lim/(x) = 0:
x—*c
For every e > 0, there exists S > 0 such that |/(x) - 0| < ewheneverO < |x - c| < 6.
Now |/(x) - 0| = |/U)| = ||/(.r)| - 0| < 6 for |x - c| < 5. Therefore, lim |/(.t)| = 0.
314 Chapter 1 Limits and Their Properties
112. (a) If lim \f{x)\ = 0, then lim [- |/(x)|] = 0.
-\f{x)\ <f{x) < l/WI
lim[-|/U)|] < lim/W < lim|/W|
0 < Urn fix) < 0
x—>c
Therefore, lim/(jc) = 0.
(b) Given \imf{x) = L:
x—*c
For every e > 0, there exists 8 > 0 such that \f{x) — L\ <e whenever 0 < \x — c\ < S.
Since \\f(x)\ ~ \L\\ < \f{x) - L\< e for \x - c\ < 8, then lim |/(;c)| = \L\.
114. True, lim jc' = (P = 0
x->0
116. False. Let f{x) =
X x i' I
3 x= 1
Then \imf(x) = 1 but /(I) # 1.
X—>1
118. False. Let/{x) = {x'^ and g{x) = x^. Then/W < g{x)
for all ;c ^ 0. But lim/(jc) = lim gix) = 0.
J— *0 x-^O
.-- ,. 1 — COS a: ,. 1 — cosx 1 + cosx
120. lim = hm
j:-*0 X j:-»0 X 1 + COS JT
= lim
1
= lim
i->o;c(l + cosj:) .t"-5bj;(l + cosjc)
sin X sin x
= lim
j->0 ;c
1 + cosj:
lim
x-»0 X
(1)(0) = 0
lim——
jr-»0 1 + cos
l]
122. fix)
secx — 1
(a) The domain of/ is all x t^ 0, ir/l + mr.
(b) 2
^J
The domain is not obvious. The hole at .jc = 0 is not
apparent.
(c) lini/W = -
(d)
sec x — I sec x — I sec j: + 1
sec^j: — 1
x^ sec a; + 1 j:^(secA: + 1)
tan^ X 1 fsin-x\ 1
Hence, lim
;c^(secAr+ 1) cos^j:\ x^ /secj:+ 1
sec ;c - 1 _ 1 /sin^A:\ 1
'o x^ x-^0 cos^ x\ 7? /sec X + 1
124. The calculator was set in degree mode, instead of radian mode.
Section 1.4 Continuity and One-Sided Limits 315
Section 1.4 Continuity and One-Sided Limits
2. (a) lim /W = -2
(b) lim f(x) = -2
(c) lim /W = -2
j:-»-2
The function is continuous at
x= -2.
ii,u 2-^ n,„ 1
1
.'ilV - 4 - iii?* x + 2-
4
4, (a) lim^/U) = 2
(b) lini fix) = 2
l->-2"
(c) lim fix) = 2
j:^-2
The function is NOT continuous at
x= -2.
6. (a) lim fix) = 0
(b) lim /U) = 2
Jc— »-l
(c) lim /U) does not exist.
jr— »— 1
The function is NOT continuous at
x = -1.
,n 1- v^- 2 ,. Ji-2 Ji+2
10. hm — = lim
j:->4- X — 4
= lim
■"- Jc-4 ^ + 2
X- A
4 U-4)(y; + 2)
lim
1
1
'-**' J~x + 1 4
,- ,. k - 2 ,. X - 2
12. lim ■' -^ = lim = 1
jr->2* X — 2 x-»2* X — 2
, ^ ,. (x + Ax)2 + (a: + Ax) - (x^ + x) ,. x~ + 2a;(Ax) + i^xf- -^x + l^-x^-x
14. lim T = lim
lim
Aj->0*
Ax
2x(Ax) + (Ax)- + Ax
Ax
= lim (2x + Ax + 1)
= 2x + 0+l=2x+l
16. lim fix) = lim (-x^ + 4x - 2) = 2
x->2* j->2*
lim fix) = lim (x^ - 4x + 6) = 2
a:->2" x-»2"
18. lim fix) = lim (1 - x) = 0
lim/(x) = 2
j->2
20. lim sec x does not exist since
j:— >Tr/2
lim sec x and lim sec x do not exist.
x->(,r/2)* Ar-.(Tr/2)"
22. lim^ ilx - W) = 2(2) -2 = 2
24. lim 1
= !-(-!) = 2
26. fix)
X + 1
has a discontinuity at x = — 1 since /(- 1) is not defined.
28./(x)= 2,
X < 1
X = 1 has discontinuity atx = 1 since /(I) = 2+ lim/(x) = 1.
2x - 1, X > 1
30. /(f) = 3 - V9 - f^ is continuous on [- 3, 3].
32. ^2) is not defined, g is continuous on [— 1, 2).
316 Chapter 1 Limits and Their Properties
34. fix) = — is continuous for all real x.
x^ + I
36. fix) = cos — is continuous for all real x.
38. fix)
— has nonremovable discontinuities atx = I and x = —I since Vim fix) and lim fix) do not exist.
j:*- — 1 jr-»l x->-l
X - 3
40. fix) = ^ _ has a nonremovable discontinuity at x ■
atx = 3 since
lim/U) = lim = -.
— 3 since lim fix) does not exist, and has a removable discontinuity
42. fix) =
x - 1
U + 2)U - 1)
has a nonremovable discontinuity atx = —2 since
lim fix) does not exist, and has a removable discontinu-
ity at j: = 1 since
lim/(;c) = lim = —.
44. fix)
\x- 3|
jc- 3
has a nonremovable discontinuity at j: = 3 since Vim fix)
does not exist. ^^
46. fix)
-2x + 3, ;c < 1
x\ x > I
has a possible discontinuity at j: = 1 .
l./(l) = 12= 1
2. A?--^^-*^ = AT- (-2x + 3) = 1
lim fix) = lim x^ = 1
3. /(I) = lim/W
lim/U) = 1
/is continuous at a: = 1, therefore, / is continuous for all real x.
\-2x, X < 2
48. fix) = 1 2 _'/! _(_ 1 o ^^ ^ possible discontinuity at j: = 2.
I ^ T'*\ I X y .-V ."^ ^
1. /(2) = -2(2) = -4
lim /W = lirn i-2x) = -4 1 iini/(;c) does not exist.
lim fix) = lim (jc^ - 4;c + 1) = -3j
x^2*-' ^ ' x^y^ ' -^
Therefore, /has a nonremovable discontinuity atx = 2.
50. /W = \ 6
|;:-3| <2
U- 31 > 2
1 < X < 5
X < 1 or j: > 5
has possible discontinuities at x = \,x = 5.
l./(l) = csc|=2 /(5) = csc^ = 2
lim/W = 2
2. lim/W = 2
3. /(I) = lim/U) /(5) = lim/(x)
x-*\ x-*5
/is continuous dXx = \ and jc = 5, therefore,/is continuous for all real x.
Section 1.4 Continuity and One-Sided Limits 317
52. fix) = tan -r- has nonremovable discontinuities at each
2k + 1 , /: is an integer.
54. fix) = 7> -\x\ has nonremovable discontinuities at each
integer L
56. lim/W = 0
lim j{x) = 0
l-»0"
/ is not continuous at x = — 4
\
/
/
,. , , ,.4 sin AT
58. hm g(x) = hm = 4
jr-»0" x-»0" X
lim gix) = lim (a — 2x) = a
x-tO* x->0'
Let a = 4.
60. lim gix) = lim
x'-a^
J:->n X — a
lim (a: + a) = 2a
Fbd a such that 2a = 8 => a = 4.
62. figix))
Jx - 1
Nonremovable discontinuity at j: = 1. Continuous for all .t > 1.
Because/ - g is not defined for x < 1, it is better to say that/ ° g is discontinuous from the right at x = 1.
64. figix)) = sin.r2
Continuous for all real x
66. hix) =
1
U + l)(x - 2)
Nonremovable discontinuity at x
— 1 and x = 2.
m
fcos .V - 1 X < 0
68. fix) = \ X
[Sx. x> 0
/(O) = 5(0) = 0
1- ft \ 1- (cosx - 1)
hm /(x) = lim = 0 -^
j:->0- .t->0- X
lim /(x) = lim (5x) = 0
a:->0* jr->0*
Therefore, lim/(x) = 0 = /(O) and/ is continuous on the entire real line, (x = 0 was the only possible discontinuity.)
x—*0
. . ^r^ '^-^,
/
70. fix) = xVx + 3
Continuous on [-3, oo]
72./U) = ^ -
VX
Continuous on (0, oc)
318 Chapter 1 Limits and Their Properties
74. f(x) =
x-1
The graph appears to be continuous on the interval
[—4, 4]. Since /(2) is not defined, we know that/has
a discontinuity at x = 2. This discontinuity is removable
so it does not show up on the graph.
76. fix) = x^ + 3a: — 2 is continuous on [0, 1].
/(0)= -2and/(l) = 2
By the Intermediate Value Theorem, fix) = 0 for at least
one value of c between 0 and 1.
— 4 TTX
78. f(x) = 1- tan — is continuous on [1, 3].
X o
/(I) = -4 + tan f < 0 and /(3) = -^ + tan ^ > 0.
o 3 o
By the Intermediate Value Theorem, /(I) = 0 for at least
one value of c between 1 and 3.
80. /(x) = x3 + 3x - 2
fix) is continuous on [0, 1].
/(0)= -2and/(l) = 2
By the Intermediate Value Theorem, /(x) = 0 for at least
one value of c between 0 and 1. Using a graphing utility,
we find that x = 0.5961.
%2. hie)= \ + e-zxaae
h is continuous on [0, 1].
MO) = 1 > 0 and /id) «
-2.67 < 0.
By the Intermediate Value Theorem, hid) = Q for at least
one value 6 between 0 and 1. Using a graphing utility, we
fmd that d = 0.4503.
84. fix) = x2 - 6x + 8
/is continuous on [0, 3].
/(O) = 8 and/(3) = - 1
-1 < 0 < 8
The Intermediate Value Theorem applies.
^2 - 6x + 8 = 0
ix - 2)(x - 4) = 0
X = 2 or X = 4
c = 2 (x = 4 is not in the interval.)
Thus,/(2) = 0.
86. fix) =
x'- + x
/is continuous on [j. 4|. The nonremovable discontinuity,
X = 1, lies outside the interval.
/
35 . ,,.. 20
-and/(4)=y
35 ^ 20
The Intermediate Value Theorem applies.
x^ + X .
x2 + X = 6x - 6
x2 - 5x + 6 = 0
(x - 2)(x - 3) = 0
X = 2 or X = 3
c = 3 (x = 2 is not in the interval.)
Thus,/(3) = 6.
Section 1.4 Continuity and One-Sided Limits 319
88. A discontinuity at j: = c is removable if you can define
(or redefine) the function at j: = c in such a way that the
new function is continuous at ;c = c. Answers will vary.
(a) fix)
(b) f(x)
|x-2|
X- 2
sin(x + 2)
X + 2
(c) fix)
1, ifx > 2
0, if-2 < X < 2
1, ifx = -2
0, ifx < -2
90. If/ and g are continuous for all real x, then so is/ + g (Theorem 1.11, part 2). However, //g might not be continuous if g(x) = 0.
For example, let/(x) = xandg(x) = x^ — 1. Then/and g are continuous for all real x, but//g is not continuous atx = ±1.
1.04, 0 < r < 2
92. C = 1 1.04 + 0.36[[r - ll, r > 2, r is not an integer
[l.04 + 0.36(f - 2), r > 2, f is an integer
Nonremovable discontinuity at each integer greater than 2.
You can also write C as
C
c
.1
4-
3--
2-
i4>
1.04, 0 < r < 2
1.04 - 0.36[2 - r], t > 2
94. Let sit) be the position function for the run up to the campsite. 5(0) = 0 (f = 0 corresponds to 8:00 a.m., 5(20) = k (distance
to campsite)). Let rit) be the position function for the run back down the mountain: Hff) = K r(\Q) = 0. Let/(r) = sit) — r(t).
When t = 0 (8:00 a.m.), /(O) = 5(0) - KO) = 0 - /t < 0.
Whenr = 10(8:10 a.m.), /(lO) = 5(10) - r(IO) > 0.
Since /(O) < 0 and /( 10) > 0, then there must be a value r in the interval [0, 10] such that /(f) = 0. If/(r) = 0, then
sit) — rit) = 0, which gives us sit) = rit). Therefore, at some time f, where 0 < t < 10, the position functions for the
run up and the run down are equal.
96. Suppose there exists x, m [a, b] such that/(xi) > 0 and there exists x, in [a, b] such that/Cx,) < 0. Then by the Intermediate
Value Theorem, /(x) must equal zero for some value of x in [x,, .r,] (or [x,. xj if x, < x,). Thus, /would have a zero in [a. b],
which is a contradiction. Therefore, /(.r) > 0 for all x in [a, b] or fix) < 0 for all x m [a, b].
98. Ifx = 0, then/(0) = 0 and Vim fix) = 0. Hence,/is
„ x->0
contmuous at x = 0.
If X T^ 0, then lim/(f) = 0 for x rational, whereas
t—*x
lim/(f) = lim kt = kx i= 0 for x irrational. Hence, /is not
continuous for all x ^ 0.
100. True
1. /(c) = L is defined.
2. lim /(x) = L exists.
Jr-*c
3. fie) = lim/(.v)
All of the conditions for continuits- are met.
320
Chapter 1 Limits and Their Properties
102. False; a rational function can be written as P(x)/Q{x}
where P and Q are polynomials of degree m and n,
respectively. It can have, at most, n discontinuities.
104. (a) s
10 IS 20 25 30
(b) There appears to be a limiting speed and a possible
cause is air resistance.
106. Let V be a real number. If > = 0, then x = O.lfy > 0, then let 0 < Xg < 7r/2 such that M = tan Xq > y (this is possible
since the tangent function increases without bound on [0, 77-/2)). By the Intermediate Value Theorem, /(x) = tan x is
continuous on [0, .^o] and 0 < >• < M, which implies that there exists x between 0 and Xq such that tan ;c = y. The argument
is similar if y < 0.
108. 1. /(c) is defined.
2. hm/{A:) = lim /(c + Ax) = /(c) exists.
[Let a: = c + Ax As;ic— >c, A;c-^0]
3. lim/(x)=/(c).
Therefore, /is continuous at JT = c.
110. Define /(jc) = /^(.t) — f^ix). Since /i and/, are continuous on [a, b], so is/
f(a) =Ua) -f,{a) > 0 and f{b) = fp) - Mb) < 0.
By the Intermediate Value Theorem, there exists c in [a, b] such that /(c) = 0.
/(c) = Mc) - /,(c) = 0 => /,(c) = /2(c)
Section 1.5 Infinite Limits
2. lim = oo
x-»-2* X + 2
lim
2-;c + 2
4. lim sec -— = oo
x->-2+ 4
lim sec -— = — oo
x-*-2- 4
6. fix
X
-3.5
-3.1
-3.01
-3.001
-2.999
-2.99
-2.9
-2.5
fix)
-1.077
-5.082
-50.08
-500.1
499.9
49.92
4.915
0.9091
\im fix) = —oo
lim fix) = oo
Section 1.5 Infinite Limits 321
8. /W = sec-
X
-3.5
-3.1
-3.01
-3.001
-2.999
-2.99
-2.9
-2.5
fix)
-3.864
-19.11
-191.0
-1910
1910
191.0
19.11
3.864
\\m f(x) = -oo
lim fix) = oo
jr-*— 3'*'
10. lim
.'-^r (x - 2)5 ^
lim 7 rrr = — oo
x-^2- ix - 2)3
Therefore, .r = 2 is a vertical asymptote.
14. No vertical asymptote since the denominator is never zero.
,^ ,■ 2+j: ,. 2+x
12. hm -r- = lim r = oo
x->o- ;t-(l - x) x-^o* x-i\ - x)
Therefore, x = 0 is a vertical asymptote.
,. 2+x
hm -:n- : = oo
/-^i-;c2(l - x)
V 2+.X
iim -57; r
— —00
Therefore, x = 1 is a vertical asymptote.
16. lim his) = —00 and lim his) = 00.
J— *-5 5— >-5*
Therefore, s = - 5 is a vertical asymptote,
lim his) = — 00 and lim his) = 00.
s—^5 s—*5 *
Therefore, .r = 5 is a vertical asymptote.
18. fix) = sec TTt
cos TTX
has vertical asymptotes at
2« + 1
-, n any integer.
22. fix]
4(.r- + .t - 6) 4U + 3)(.t - 2)
20. gix) =
(l/2).t3 - .t- - 4.t _ 1 a:(.v- - 2x - 8)
3x- - 6.t - 24 ~ 6 j:- - 2jc - 8
-JC,
;c?t -2,4
No vertical asymptotes. The graph has holes at x
and x = 4.
T,xi= -3.2
.r(x3 - Ix' - 9x + 18) xix - 2){x~ - 9) xix - 3)'
Vertical asymptotes at .r = 0 and .v = 3. The graph has holes at jt = - 3 and x = 2.
24. hix) =
(x + 2)(.t - 2)
f(f-2)
;c3 + 2x- + .r + 2 U + 2)(.t2 + 1)
has no vertical asymptote since
lim hix) = lim ^ = — -r.
x-»-2 x-,-2X- + 1 5
f =t 2
^ ' (f - 2)(r + 2)it- + 4) (r + 2)(f- + 4)'
Vertical asymptote at r = — 2. The graph has a hole at
322 Chapter 1 Limits and Their Properties
,„. tan 0 sin S , . ,
28. g(0) = — — = has vertical asymptotes at
V 6 cos 6
30. lim - — ^V^ = lim U - 7) =
jt->-l X + \ Ar-»-l
-8
(2m + 1)77 77
6 = = — + nv, n any integer.
There is no vertical asymptote at 0 = 0 since
tan e
hm— — = 1.
e->o Q
Removable discontinuity at x = — 1
,, ,. sm(:ic + 1)
32. lim — —^ = 1
jr->-l X + I
Removable discontinuity at ,3
x= -1
34. hm = -00
x->l* I — X
36. lim
x-*4- x~ + \6 2
38. lim
6jc2 + ;c - 1
lim
3jc- 1 5
x->-(i/2)* Ax^ - 4x - i j->-(i/2)+ 2x - 3
40. lim ^-T^ = \
42. lim x^ - - = 00
x^o- \ X
44. lim
x-tMl)-^ cos j:
46. lim i^i^ = lim [(x + 2)tan ;c] = 0
;t->0 cot X x-»0
48. lim x^ tan ttj: = 00 and lim x^ tan ttj:
Therefore, lim x^ tan ttx does not exist.
a:->(l/2)
50. /U) =
X^- 1
X^ + X + I
lim /U) = lim U - 1) = 0
x-*l x—*l
52. fix) = sec -
lim fix) = — 00
I
J.
Pi
in
54. The line x = c is a vertical asymptote if the graph of/
approaches ±co asx approaches c.
56. No. For example, fix) =
vertical asymptote.
•has no
58. P
lim — = fe(oo) = 00 (Tn this case we know that k > 0.)
V-.0* V
Section 1.5 Infinite Limits 323
,77 IOOtt „ ,
60. (a) r = SOtt sec- - = -^— ft/sec
6 3
(b) r = 50-n- sec^ ^ = 200-77 ft/sec
(c) lim [5077 sec- 6] = co
62. m =
Vl - (vVc^)
lim m = lim , ° ^^
64.
(a) Average speed =
Total distance
Total time
50 =
2d
id/x) + (d/y)
50 =
2xy
y + x
50}-
f 50^ =
50,v =
50.r =
2xy
2xy - SOy
2yU - 25)
X
25;t
- 25
y
Domain:
X > 25
66.
(a)
A=^bh
-¥'
= ^(10)(10tar
50 tan e - 50 0
Domain: 0
X
30
40
50
60
y
150
66.667
50
42.857
(b)
(c) lim — = oo
x^25* .X — 25
As X gets close to 25 mph, y becomes larger and larger.
(b)
0
0.3
0.6
0.9
1.2
1.5
fie)
0.47
4.21
18.0
68.6
630.1
(c)
(d) lim A = oo
e->-ir/2"
68. False; for instance, let
x^- 1
f(x)
X - \
70. True
The graph of/has a hole at (1, 2), not a vertical
asymptote.
72. Let/(.r) = "J and gix) = -j, and c = 0.
lim -^ = oo and lim —7 = 00, but
x-)0 AT x->0 X
1 1
jt->o v.-r j:^/ j-»o
i(^) =
-00 ^t 0.
?{v)
74. Given lim f{x) - 00. let ? (.v) = 1 . then lim ^^-r- = 0
X -tr .1 ->c J{X)
by Theorem 1.15.
324 Chapter 1 Limits and Their Properties
Review Exercises for Chapter 1
2. Precalculus. L = J{9 - If + (3 - 1)^ == 8.25
4.
X
-0.1
-0.01
-0.001
0.001
0.01
0.1
fix)
1.432
1.416
1.414
1.414
1.413
1.397
Urn fix) = 1.414
6. six
2x
x-2
(a) lim gix) does not exist.
jr-»2
(b) lim^W =0
x—*0
8. lim 7^ = 79 = 3.
jr->9
Let 6 > 0 be given. We need
I V5 — 3| < 6 => \~/x + 3|| VJ "■ 3| < e| V5 + 3 1
\x - 9\ < e\Vx + 3\
Assuming 4 < x < 16, you can choose 6 = 5e.
Hence, for 0 < |j^ - 9| < S = 5e, you have
|;c - 9| < 56 < \Vx + 3|e
I V5 - 3| < 6
l/W -L\<e
10. Um 9 = 9. Let e > 0 be given. 5 can be any positive
number. Hence, for 0 < |j: — 5 1 < 5, you have
|9 - 9| < e
\f(x) - L| < e
12. Iim3|>' - ll = 3|4- ll = 9
14. lim
l->3 f — 3 r-»3
= lim (r + 3) = 6
,^ ,. V4 + ;c - 2 ,. ^4 + X - 2 ^4 + ;f + 2
16. lim = lim • , — r
j:-»o j: j^o X ^4 + X + 2
1
lim - , = -
^^0 V4 + X + 2 4
18. lim-
j->0
(i/Vi + s)- ] ^ ,.^ [(i/Vi +5) - 1 _ (i/Vi + j) + 1
^ (l/Vl + s)+ I.
Um
^^^^ [1/(1 + .)] - 1 ^ ^.^^^ -1
^-^0 5[(i/v'T+7)+ 1] ^™(i +4(i/yrT7)+ 1]
20. lim
x^-2 a:-' +
= lim
(;t + 2)ix - 2)
2 (x + 2)(a:'- - 2;c + 4)
x-2
4x 4(7r/4)
22. um = — ; — = TT
j>-^(jr/4) tanx 1
= lim , --, — y
x-.-2X^ - 2x + 4
4_
12
Review Exercises for Chapter ] 325
24. lim
A:t->0
cos(7r + Ax) + 1
Ax
lim
cos TT COS Ajf — sin TT sin Ajc + 1
Ax
(cos Ax -
Ax
= -0 - (0)(1) = 0
= lim
Ax-fO
D] ,. r . sinAx]
— — lim sm 17 — : —
J Ax-.0|_ Ax J
26. lim [/(x) + 2g{x)] = -I + 2(f) = ^_
28. f(x)
X- 1
(a)
X
1.1
1.01
1.001
1.0001
fix)
-0.3228
-0.3322
-0.3332
-0.3333
lim
1* X - 1
= -0.333 (Actual limit is -|.)
, 1-3/J ,. 1-3/^ l+V^+ilGf
lim — = lim — • -p — } Jy
x-,r X - 1 x-,V X - 1 1 + 3/^ + (i/x}^
-™* (x - l)[l + 4/1 + (^']
= lim
1
'* 1 + ^/5 + (^'
(b)
30. j(f) = 0 => -4.9r- + 200 = 0 => f= » 40.816 ^ t = 6.39 sec
When t = 6.39, the velocity is approximately
lim^^^^-=^ = lim-4.9(^ + r)
t-*a a — t t—*a
= lim -4.9(6.39 + 6.39) = -62.6m/sec.
32. lim |.ic - IJ does not exist. The graph jumps from 2 to 3
X— *4
at X = 4.
34. lim g{x) =1 + 1=2.
36. lim f{s) = 2
38. fix)
"ix- — x — 1
lo,
X - 1
X ^ 1
x= 1
3r^ - t - 2
lim/(.x) = lim — '—, — -
.t-»l .t-»l X — 1
= lim (3x + 2) = 5 ^ 0
Removable discontinuity at x = 1
Continuous on (-co, 1) u (1, cx>)
326 Chapter 1 Limits and Their Properties
40. fix) =
5 - X, X <2
2jc - 3, X > 2
lim (5 — jc) = 3
lim (2a: - 3) = 1
Nonremovable discontinuity at x = 2
Continuous on (- oo, 2) U (2, oo)
n ^/l
lim
1 + - = oo
Domain: (-oo, - 1], (0, oo)
Nonremovable discontinuity at x = 0
Continuous on (— oo, — 1] U (0, oo)
44. f{x) =
lim
x+ 1
2x + 2
X + 1 1
:r-.-l 2{X +1) 2
Removable discontinuity atx = —I
Continuous on (— oo, — 1) U (- 1, oo)
46. fix) = tan 2x
Nonremovable discontinuities when
(2n + l)7r
Continuous on
/(2w - l)7r (2n + Dtt'
V 4 ' 4
for all integers n.
48. lim (x + 1) = 2
lim (x + 1) = 4
jr-»3"
Find ii and c so that lim (x^ + to + c) = 2 and lim (x^ + to + c) = 4.
Consequently we get \ + b + c = 1 and 9 + 3i + c = 4.
Solving simultaneously, Z? = - 3 and c = 4.
50. C = 9.80 + 2.50[-[-xI - 1], x > 0
= 9.80 - 2.50DI-xl + 1]
C has a nonremovable discontinuity at each integer.
52. fix) = V(x - l)x
(a) Domain: (-00, 0]u [1, oo)
(b) lim_/(x) = 0
(c) lim^/(x) = 0
54. hix) =
4x
4-x2
Vertical asymptotes at x = 2 and x = — 2
56. fix) = CSC TTX
Vertical asymptote at every integer k
58. lim
(1/2)* 2x - 1
60. lim -J = lim
-= -i
-"1- x" - 1 " x^-\- i^ + l)(x - 1) ~ 4
^, ,. x^ - 2x + 1
62. hm ; = 00
j:-»-1* X + 1
64. lim
^l/^F^^A
,, ,. secx
66. lim — ■ — = 00
JT-'O* X
68. lim
j:-»0" X
Problem Solving for Chapter 1 327
70. f(x) =
tanlr
X
-0.1
-0.01
-0.001
0.001
0.01
0.1
fix)
2.0271
2.0003
2.0000
2.0000
2.0003
2.0271
(a)
, . tan 2;c
hm = 2
Ji-»0 X
(b) Yes, define
/(.
rtai
tan 2x , X i' 0
x = 0
Now/(x) is continuous atjc = 0.
Problem Solving for Chapter 1
2. (a) Area APAO = ^bh = |(1)W = |
Area APBO = ^bh = |(1)().) = f = f
^, , , Area APBO a:72
(b) a(x) = , „. ^ = — 7— = X
' Area APAO .x:/2
X
4
2
1
0.1
0.01
Area APAO
2
1
1/2
1/20
1/200
Area APBO
8
T
"
1/2
1/200
1/20,000
a{x)
4
2
1
1/10
1/100
(c) lim a(x) = lim a: = 0
l->0* jr->0*
4. (a) Slope =
4-0 ^ 4
3 -0 ~ 3
3 3
(b) Slope = - - Tangent line: y — 4 = — ^x — 3)
3 25
-4^ + T
(c) Let e = (x, y) = (.v, V25 - x^)
725 - .t^ - 4
X- 3
,. V25 -X--4 V25 - .r- + 4
(d) lim m^ = lim r • — ,
.t_>3 -^ _j^3 .V - 3 V25 - .r= + 4
25 - .v= - 16
= lim
3 (x - 3)(V25 - x^ + 4)
,. (3 - x)(3 + x)
-'^3 (.V - 3)(V25 - .r= + 4)
,,„, -(3+.T) -6 _ 3
— Mm — , — 7 — — —
^^3 V25 - x^ + 4 4 + 4 4
This is the slope of the tangent line at P.
J a + bx - J3 Va + to - V3 Va + to + V3
■v Va + to + V3
(a + to) - 3
A:(Va + to + v^
Letting a = 3 simplifies the numerator.
Thus,
to
,. vT+to- V3 ,.
lim = lim / . = T^r
•"^^o .x: -<^o.v(V3 + to- + 73)
■*o V3 + to- + v/3'
Setting ■
73 + V3
Thus, 0 = 3 and ii = 6
= >/3, you obtain b = 6.
8. lim fix) = lim (a= - 2) = a= - 2
x-tO- .r->0-
o-t / tan .V
lim fix) = lim = a because lim = 1
j:->o' .t-i-o*tanj: V .>->o x
Thus,
a- — 2 = a
a~- a-2 = 0
{a - 2)ia +0 = 0
a= -1.2
328 Chapter I Limits and Their Properties
10.
-• -2--
o« -
0>
o«
(a) /(i) = W = 4
/{3) = i = 0
/(I) = lU = 1
(b) lim fix) = 1
lim /W = 0
lim/(jc) = -oo
x—*0~
lim fix) = oo
(c) /is continuous for all
real numbers except
x = 0,±l, ±i±i. .
12. (a) V- =
192,000
2 192.000 , 2 .-
= v2 - Vq^ + 48
192,000
'" V - Vq- + 48
192,000
lim r = — r
v->o 48 - Vf,^
(b)
Let Vq = V48 = 4 V3 feet/ sec.
, 1920
1920
+ Vq^ - 2.17
Vo^ + 2.17
14. Let a T^ 0 and let e > 0 be given. There exists 8, > 0
such that if 0 < l;c - 0| < S, then \fix) - i| < e.
Let 8 = 5J\a\. Then for 0 < |x - 0| < S = 8J\a\,
you have
lax < 8,
l/M - L| < 6.
As a counterexample, let /U) =
Then Wmfix) = 1 = L,
but \m\fiax) = liin/(0) = 2.
X7t 0
x = 0'
1920
v2 - vo^ + 2.17
1920
hm r = -—-z ^
v^o 2.17 - v^
(c)
Let Vq = V2.17 mi/sec (= 1.47 mi/sec).
10,600
Vo2 + 6.99
lim r =
10,600
6.99 - v„
Let Vq = V6.99 = 2.64 mi/sec.
Since this is smaller than the escape velocity for earth,
the mass is less.
CHAPTER 2
Differentiation
Section 2.1 The Derivative and the Tangent Line Problem . . 330
Section 2.2 Basic Differentiation Rules and Rates of Change 338
Section 2.3 The Product and Quotient Rules and
Higher-Order Derivatives 344
Section 2.4 The Chain Rule 350
Section 2.5 Implicit Differentiation 356
Section 2.6 Related Rates 361
Review Exercises 367
Problem Solving 373
CHAPTER 2
Differentiation
Section 2.1 The Derivative and the Tangent Line Problem
Solutions to Even-Numbered Exercises
2. (a) m
1
4
(b) m = 1
/(4)-/(3) 5 - 4.75
4-3
1
0.25
Th..e /(4)-/(l), /(4)-/(3)
^"«- 4-1 > 4-3
(b) The slope of the tangent line at (1, 2) equals /'(I).
This slope is steeper than the slope of the line
through (1, 2) and (4, 5). Thus,
«
-»
^<;' -('«</•(.).
6. g{x) = -x+ I IS SL line. Slope = -
8.
Slop. . ,0.1). lim * + '^'-'<»
Aa^o Aj:
,. 5 - (2 + Axy - 1
= hm ,
Ajr-»0 Ax
,. 5 - 4 - 4(Ax) - (Ax)' - 1
= lim 1
Ai->0 Ax
= lim (-4 - Ajc) = -4
Ax^O
10. Slope at ( 2,7)-lim'^^-2 + ^;)-'^(-
-2)
12.
8(x} = -5
,. (-2 + At)2 + 3 -
= lim 1
Ar—O Af
_7
^'« = il5o^^^^^^
B ,. 4 - 4(Ar) + (Ar)2
= lim T
Ar->0 .Af
- 4
= lim-^-/-^)
Ar->0 Ax
= lim (-4 + Ar) = -4
= lim -^ = 0
Ai:->0 Ax
14. fix) = 3a: + 2
16
. fix) = 9 - |;c
/'W=lim^(^^t^)-^«
Ar-»0 Ax
/'(x)=lim^(^ + ^)-/«
■^ Ax->0 Ax
- lim '■^^'^ + ^) + 2] - [3x + 2]
Ar^O AjC
,.^^^ [9 - (1/2)U + Ax)] - [9 - (l/2);c]
At-»0 Ax
,. 3Ax
= lim -; —
Ai->0 Ax
= lim -- = --
Ac^o V 2/ 2
= lim 3 = 3
AX-.0
330
Section 2. 1 The Derivative and the Tangent Line Problem 331
18. fix) =l-x^
nx + ^x)-f{x)
^(^^ = i,'?o K^
,. [1 - {x + Axn - [1 - x"]
= lim :
Aj-»o Ax
,. I - x^ - IxAx - [Axf - \ + }?
= lim ;
Aa:->o Hoc
-2xAx - iiyx)' ,. , „ .X
= hm : = lim ( — 2x — Ajc) = —2x
20. f{x) =x^ +x^
■' Ai-»0 Ax
,. [U + A.t)3 + {x + Ax)2] - [x^ + .t2]
= lim :
Ai-»0 Ax
= lim
■T^ + 3.T^A.x + 3x(Ax)- + (A-x)^ + .^ + ltA.r + (A.x)- - x-' - x^
Ax
,. S.x^A-x + 3x(Ax)2 + (A.x)' + IxAx + (A.x)-
= lim -.
A^-»0 Ax
= lim (3x^ + 3xAx + (Ax)^ + Ix + (Ax)) = 3x^ + Zx
22. fix) = ^
fix) = lim
= lim
A;[->0
/(x + Ax)-/(x)
A.X
1
1
(x + Axf x"
Ax
,. X- - (x + Ax)^
~ lim -: — ; . ., -,
^x^o Ax{x + Ax) XT
— lim
-2xA.x - (Ax)^
A^^o A.x(x + Ax)V
,. -2x - Ax
lim 7 — , K \-> ■>
tix-^o (x + Ax)-x"
-Ix
X*
_2_
x3
24. /(x) = -^
vx
/'(x) = hm
Ax— »0
/(x + A.x) - /(x)
Ax-
= lim
Ax-»0
Jx + A.X J~x
Ax
= lim
4^ - 4Vx + A.X ( Jx + Jx + Ax
^-^0 AxVxVx + A.V Vv'x + Jx + Ax
Ax - 4(x + A-x)
= lim
A^^o Axv^Vx + A.x(v''x + Vx + A.x)
-^^o VxVx + A.x(^/x + Vx + Axr)
-4 _ -2
v/xVx(Vx + v'x) xvx
332 Chapter 2 Differentiation
26. (a) /(*) = ^2 + 2;c + 1
f{x + ^x)-fix)
f{x) = lim
,. [{x + Ax)^ + 2{x + Ax) + l]-W + 2x+ I]
— jjjj,
Ar^O Ax
= lim
Ax-»0
2xAx + (Axf + 2Ax
Ax
= lim (2;c + A;c + 2) = 2;c + 2
Aji->0
At (-3, 4), the slope of the tangent line is w = 2(- 3) + 2 = —4.
The equation of the tangent line is
y - 4= -4(x + 3)
y = -Ax - 8.
28. (a) fix) = x3 + 1
f'(x) = lim
fix + Ax) -fix)
Ax
= lim
Ai-^O
lim
[U + AxY + 1] - (;c3 + 1)
Ax
jc^ + 3j:^(A;c) + 3.t(Aj:)^ + (Ax)^ + 1 - x^ - 1
Ax
= lim [3x2 + 3^(^) + (^)2] = 3^2
Ar-»0
At (1, 2), the slope of the tangent line is m = 3(1)^ = 3.
The equation of the tangent line is
:y - 2 = 3(x - 1)
y = 3x - 1.
(b)
(-3,4rt
/
\
(b)
30. (a) fix) = Vx - 1
■' Ax->0 Ax
,. Vx + Ax - 1 - Vx - 1 / Vx + Ax - 1 + Vx - 1
= iim : • — . = ,
t^^o Ax \Jx + Ax - 1 + Vx - 1
(x + Ax - 1) - (x - 1)
'^^o Ax(Vx + Ax -T + Vx - l)
1 1
= lim — , , - ,
Aat^ Vx + Ax - 1 + Vx - 1 2Vx - 1
At (5, 2), the slope of the tangent line is
1
1
til — , — .
2V5 - 1 4
The equation of the tangent line is
(b)
y - 2 = -(x - 5)
Section 2. 1 The Derivative and the Tangent Line Problem 333
32. (a) /W = ^^
(b)
f'{x) = lim
/(;c + M-/U)
Ax
lim
Ax-^O
lim
X + Ax + 1 -t + 1
Ax
(x + 1) - (x + Ax + 1)
Ai^o Ax(x + zlx + l)(x + 1)
Ax-^O (x + Ax + l)(x + 1)
^ 1
At (0, 1), the slope of the tangent line is
The equation of the tangent line is y = — x + 1.
34. Using the limit definition of derivative, /'(x) = 3x'. Since
the slope of the given line is 3, we have
3x- = 3
x2 = l=>x = ±1.
Therefore, at the points (1, 3) and (- 1, 1) the tangent
lines are parallel to 3x - y - 4 = 0. These lines have
equations
.V - 3 = 3(x - 1) and y - 1 = 3(x + 1)
y — 3x y = 3x + 4
(0. rp
"^
\
36. Using the limit defmition of derivative, fix) =
-1
2(x - D''^-
Since the slope of the given line is --, we have
1
1
2(x - l)'-^ 2
1 = (x - D^^^'
1 = X - 1 => X = 2
At the point (2, 1), the tangent line is parallel to
X + 2y + 7 = 0. The equation of the tangent line is
y-l = --ix-2)
y = -^x + 2
38. h(- 1) = 4 because the tangent line passes through (- 1, 4) 40. /(x) = x- => /'(x) = 2x (d)
6-4 2 1
hX-\) =
3-(-l) 4 2
42. /' does not exist at x = 0. Matches (c)
44.
Answers will var>'.
Sample answer: y = x
334 Chapter 2 Differentiation
4^ . ^ V r fix + 2Ax)-f(x) .. f(x + ^x)-f{x)
46. (a) Yes. lim r-- = lim -. = / (x)
Ax^O 2Dx Ax->0 Ax
(b) No. The numerator does not approach zero.
f(x + Ax)-f{x-Ax) fix + Ax) - fix) - fix - M + fix)
(c) Yes. lim — = lim —
aj:-»o 2Aa; Ai^o 2Aj:
= lim
Al->0
fix + Ax) -fix) fix -Ax) -fix)
lAoi 2i-Ax)
= \rix)+\f'ix)=f'ix)
(d)Yes. lim^^^-^y-^^^^^/'M
AX-.0 AjC
48. Let (jcq, >(,) be a point of tangency on the graph of/. By the limit definition for the derivative, fix) = 2x. The slope of the line
through (1, - 3) and (xg, yg) equals the derivative of /at x^:
%
2Xn
-3 -Jo = (1 - JCo)2xo
3 ;cq"- 2Xg
- 2xo - 3 = 0
^r 2
■^^0
(Xg - 3)(A;g + 1) = 0 => Xg = 3, - 1
Therefore, the points of tangency are (3, 9) and (- 1, 1), and the corresponding slopes are 6 and - 2. The equations of
the tangent lines are
y + 3 = 6U - 1) y + 3 = -2(x - 1)
y = 6x — 9 y = —2x — I
50. (a) fix) = x2
fix)
lim
Ax-fO
= lim
= lim
Ajr->0
= lim
Aj:-»0
fix + Ax) -fix) •
Ax
ix + Ax)^ - x^
Ax
x^ + 2xiAx) + iAx)^ - x^
Ax
Ar(2jc + Ax)
Ax
= lim (2x + Ax) = 2x
Aj->0
At X = - !,/'(- 1) = — 2 and the tangent line is
y - 1 = -2(x +1) or y = -2.x - 1.
At X = 0,/'(0) = 0 and the tangent line is y = 0.
Atx = 1,/'{1) = 2 and the tangent line is y = 2x
\
/
\
\
For this function, the slopes of the tangent lines are
always distinct for different values of x.
(b) g'ix) = lim
= lim
Ax^O
= lim
Ai->0
= lim
Ax->0
g(x + Ax) - g(x)
Ax
(x + Ax)^ - x'
Ax
x^ + 3x^(Ax) + 3x(Ax)^ + (Ax)^ - x^
Ax
Ax(3x^ + 3x(Ax) + (Ax)^)
Ax
= lim (3x2 + 2xiAx) + (Ax)^) = 3x^
Ax->0
At X = - 1, g'i— 1) = 3 and the tangent line is
y + 1 = 3(x + 1) or y = 3x + 2.
At X = 0, g'iO) = 0 and the tangent line is y = 0.
At X = 1, g'il) = 3 and the tangent line is
y - 1 = 3(x - 1) or y = 3x - 2.
For this function, the slopes of the tangent lines are
sometimes the same.
Section 2. J The Derivative and the Tangent Line Problem 335
52. fix) = kx^
By the limit definition of the derivative we havsf'ix) = x.
X
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
m
2
1.125
0.5
0.125
0
0.125
0.5
1.125
2
fix)
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
54. g(x)
fix + Om)-f(x)
0.01
= {3Vx + 0.01 - 3v/3)lOO
56. /(2) = ^23) = 2,/(2.1) = 2.31525
/'(2) » "'f^ ~ ^ = 3.1525 [Exact:/'(2) = 3]
The graph of g(x) is approximately the graph of/'(x)-
58. fix) = --l>x and/'U) = -x~ - l
60. fix) = ;c
/(2 + Ax)-/(2) (2 + A^)+^^^
5a^ W ^^ U - 2) + /(2) = -^ ix - 2) + -
2(2 + M- + 2-5(2 + M. 5 (2A.X + 3) _ ,. , 5
2(2 + Ax) A.X ^ ''2 2(2 + Av) '2
(a)Ajt= 1: 5^, = -(jf- 2) +- = -J: +
5 5 5
2 = 6" -"6
Ar = 0.5: 5^ = 7(x - 2) +
5 4
:.v + —
16, ~~ 5 ^ 2^ 41^
2 ~ 21-^ "^42
(b) As Aat— >0, the line approaches the tangent line to/at (2. ?).
/Vx = 0.1:5^ = -(.v-2) + - = -. + 42
r X
Y
)^/r\
62. ^(.r) = .rCt - 1) = x^ - .t, c = 1
»,x ,. six) - g(l) X- - .t - 0 ,. .v(x - 1)
g (1) = lim ; — = lim ; — = lim —
J:->1 X - 1 .t-.l X - 1 v-»: X - 1
= lim X = 1
336 Chapter 2 Differentiation
64. f(x) = j^ + Zx,c= I
r,„. ,. /(-t)-/(l) ,. x^ + 2x-Z ,. (x - 1)U^ + ^ + 3) ,. .2^ ■ ,. -
/ (1) = lim ~ = lim ; = lim ; = lim (x^ + x + ?>) = 5
x-,\ X - \ x->\ X - \ J->1 X - \ jr-»l
66. f(x) = -, c = 3
X
x->3 X — 3 jr->3 jc - 3 j:->3 3a: X — 1> -r-»3 V 3;c/ 9
68. gW = (x + 3)'''3,c= -3
g'(-3) = lim ^W^/ll^= lim^-^^n'°= l™ ^^
Does not exist.
70. f(x) = |j; - 4|, c = 4
^,(4) = ,im^W^= lim ^^1^ = lim ^^^
^4 X - A x^i X — 4 x^4 X - 4
Does not exist.
72. /(.t) is differentiable everywhere except at x = ±3. (Sharp turns in the graph.)
74. f{x) is differentiable everywhere except atx = 1. (Discontinuity)
76. fix) is differentiable everywhere except at x = 0. (Sharp turn in the graph)
78. fix) is differentiable everywhere except at jc = ±2. (Discontinuities)
80. fix) is differentiable everywhere except at x = 1. (Discontinuity)
82. fix) = yp^
The derivative from the left does not exist because
,. /W-/(l) ,. Vl - x2 - 0 ,. Vl -x2 Vl -;c2 ,. 1 + r .,, ■ ,
lim ; — = lim ; = lim ; — • . = lim , = — oo. (Vertical tangent)
x-^i' x — I x->i- X - \ x->i- j: — 1 VI - Jc^ -t^'^ J\ — x?-
The limit from the right does not exist since/is undefined for jt > 1. Therefore,/is not differentiable at x = 1.
X, X < \
x^, X > I
84. /U) ,,
The derivative from the left is
,. /U)-/(l) ,. x-1 ,. , ,
lim ^ = lim = lim 1 = 1.
x->i' ;c — 1 x^\~ X — I x^i-
The derivative from the right is
fix) -/(I) ,. jc^ - 1 ,. , ,, ^
hm = -^-^-^^ — f!^ = lim = lim ix + \) = 2.
jr-»r X — 1 Jr-»1* X — I Ar-»r
These one-sided limits are not equal. Therefore, / is not differentiable at j: = 1.
Section!.] The Derivative and the Tangent Line Problem 337
86. Note that/ is continuous at x = 2.f(x) =
The derivative from the left is
lim
3* + 1, .r < 2
V^, X > 2
r-»2- X - 2 x->2- X — 2 x-y2- X — 2 2
The derivative from the right is
,. fix) - m ,. v/5-2 72^ + 2
lim — = lim — • — 7=
x-^2* x-2 x->2^ X -2 J2x + 2
1- 2.V-4 2ix - 2) 2 1
- iiJ?* (.-2)(72^ + 2) = .'i'?M.-2)(72^ + 2)= ii-^^T^T^ = i
The one-sided limits are equal. Therefore, / is differentiable at j: = 2. (/'(2) = 5)
88. (a) fix) = x^ and/'U) = 2x
(b) gix) = ;c' and g'U) = 3.r^
(c) The derivative is a polynomial of degree 1 less than the original function. If hix) = x", then /('(.t) = nx" '.
(d) If/W = A then/'U) = lim ^^ + ^ - /^
,. U + A.r)-* - ;c^
= Imi :
Aj->o Ax
lim
Ax->0
= lim
Ax-*0
.r* + 4.r3(Aj:) + 6xHAx)- + 4.T(A.r)^ + (A.r)^ - x*
Xx
Ax(4.t3 + ex'jAx) + 4x(A.r)^ + (Ax)^)
A.V
= lim (4x3 + 6.i2(A.x) + 4.r(Ax)= + (Ax)^) = 4x^
Ajt-»0
Hence, if/(x) = x^, then/'(x) = 4x3 which is consistent with the conjecture. However, this is not a proof, since you must
verify the conjecture for all integer values of n,n> 2.
90. False, y = |x - 2| is continuous at x = 2, but is not differentiable at x = 2. (Sharp turn in the graph)
92. True — see Theorem 2. 1
94.
As you zoom in, the graph of v, = x- + 1 appears to be locally the graph of a horizontal line, whereas the graph of
J": ~ kl + 1 always has a sharp comer at (0, 1). v, is not differentiable at (0, 1).
338 Chapter 2 Differentiation
Section 2.2 Basic Differentiation Rules and Rates of Change
2. (a) y = jc-'/2
(b) y = X-'
y'--hr
3/2
y'= -AT
y'W = -5
y'W=-i
4. fix) = -2
6. >' = ;c«
fix) = 0
>-'= 8;c''
(c) y = x-'/2
(d) y = x-2
y'=-!^-
5/2
y'= -2x-3
^-'d) = -1
y'W = -2
8. , = ! = .-
10. :y = 4^ = x^'
y' = Sx"
V' = -X-3/'* = ^—
-*^ 4 4xV4
12. gW = 3;c - 1
14. y =
r^ + 2? - 3
16. y
= 8-x3
18. fix) = 2x3 -
g'(x) = 3
y' =
2f + 2
y'
= -3x2
fix) = 6x2 _
20. g(r) = IT cos f
22. .v =
5 + sinx
24. V
— /„ M + 2 cos X — „x 3 + 2 COS X
(2x)3 8
g'{t) = -TTsinf
>'' =
COSJC
y'
= |(-3)x-
"^ - 2 sin X = — -J — 2 sin X
Function
Rewrite
Derivative
Simplifv
''■ y = 3^ .
2 -2
>- = JX 2
r--\.-'
4
^ - 3x3
^'■y^iZy
y--'-r-'
27r
y - 9x3
30. . = ^3
y = 4;c3
y'=\2£-
y' = 12«2
32. /W = 3-|.(f,2)
34. y =
= 3x3-6, (2, 18)
36. /(x) = 3(5 - x)\ (5, 0)
m-h
• y'-
-- 9x-
= 3x2 _ 30;t + 75
3''(2) =
= 36
fix) = 6x - 30
. /'(M
/'(5) = 0
38. g{i) = 2 + 3 cos f, (rr.
-1)
40. /U) =
x^-3x- 3x-2
42. fix) = X + x-2
g'W = -3sinf
fix) =
2x - 3 + 6x-3
/'(x) = 1 - 2x-3
gV) = 0
=
2X-3+4
X?
--?
, , , 2x2 _ 3^ + J
44. hix) = = 2x - 3 + X-'
, . , - 1 2x2- 1
ft (x) = 2 r =
X^ x2
46. y = 3x(6x - 5x2) = ig^2 _ 15^^
y' = 36x - 45x2
48. fix) =Vx+Vx = xi/3 + xi/5
50. fit) = r2/3 - ri/3 + 4
/'W4x-2/3+ix-/3._l_^_^
2 1 2 1
/W 3^ 3^ 3,1/3 3^/3
Section 2.2 Basic Differentiation Rules and Rates of Change 339
2
52. fix) = ^7= + 3 cos X = 2jt-'/5 + 3 cos Jt
— 2 -2
/'W = — Jc""*/^ - 3 sm;c = ^^ - 3 sinjc
54. (a) y = x> + X
y' = 3jc2 + 1
At (-1,-2): y'= 3{-\Y +1=4.
Tangent line: y + 2 = 4U + 1)
4jc - y + 2 = 0
(b)
/
'J
/i
56. (a) y = (x2 + 2x)(;c + 1)
= x? + 2,x- + 2x
y' = Zx- + 6x + 2
At (1,6): y' = 3{\)- + 6(1) + 2 = 11.
Tangent line: y - 6 = ll(.x - 1)
0= \\x- y - 5
(b)
T^^
58. y = ;<^ + .X
;y'= 3;c2 + 1 > 0 for all x.
Therefore, there are no horizontal tangents.
60. v = ;c- + 1
>>' = Iv = 0 => .I = 0
At;c = O.y = 1.
Horizontal tangent: (0. 1)
62. y = JTix + 2 cos at, Q < x < Itt
y' = ^ - ls\r\x = Q
J3
TT 2tT
x = yor-
TT 73Tr+ 3
At :ic = — , >> =
277 2v^Tr - 3
At j: = — , y = ^ .
Horizontal tangents:
TT JItt + 3\ /27T 2x/3Tr - 3
3"
3 '
64. k ~ X- = — 4jc + 7 Equate functions
— 2r= —4 Equate derivatives
Hence, a: = 2 anclA:-4=-8 + 7=>it = 3
66. kjl(. = j: + 4 Equate functions
Equate derivatives
2j~x
Hence, k = 2>/x and
(27x)v^ = x + 4^>lt = .r + 4=>.v = 4=i.fc = 4
68. The graph of a function/
such that/' > 0 for all x and
the rate of change the function
is decreasing (i.e./" < 0)
would, in general, look like the
graph at the right.
/
(P?^
yi
•7
-X^
340 Chapter 2 Differentiation
70. gix) = -5fix) => g'ix) = -Sf'ix)
72.
If/ is quadratic, then its derivative is a linear function.
fix) = ay?- + bx + c
fix) = 2ax + b
74. m, is the slope of the line tangent to >> = x. mj is the slope of the line tangent toy = l/x. Since
y = x
y' = 1 => m^ = \ andy =
^ =1^
The points of intersection ofy = x and y = l/x are
1
jc = —
X
±\.
Atjc = ±l,m2=— 1. Since wij = — \/m^, these tangent lines are perpendicular at the points intersection.
76. fix) = -, (5, 0)
fix)
_1_
x^
x^ 5 — X
- 10 + 2x = -x^y
.V2
■ 10 + 2x = -X-
•10 + 2x = -2x
4x = 10
x = -^,y
The point (j, j) is on the graph of/. The slope of the
tangent line is/'^j) = -js-
Tangent line: y ~ 1
'25 ^ 2
25y - 20 = -8x + 20
8x + 25>' - 40 = 0
78. /'(4) = 1
dy/dx=l
^
Section 2.2 Basic Differentiation Rules and Rates of Change 341
80. (a) Nearby point: (1.0073138, 1.0221024)
1.0221024 - 1
Secant line: y - I
1.0073138 - 1
y = 3.022(x - 1) + 1
(Answers will vary.)
<;c-l)
/(l.l)
f
(b) fix) = 3x^
Tix) = 3{x- I) + \ =2x -2
(c) The accuracy worsens at you move away from (1, 1).
2
/■"
'f
/
(d)
Ax
-3
-2
-1
-0.5
-0.1
0
0.1
0.5
1
2
3
fix)
-8
-1
0
0.125
0.729
1
1.331
3.375
8
27
64
T{x)
-8
-5
-2
-0.5
0.7
1
1.3
2.5
4
7
10
The accuracy decreases more rapidly than in Exercise 59 because y = x^ is less "linear" than y = x''^.
82. True. If/W = g{x) + c, then/'W = gXx) + 0 = g'{x).
86. False. If f{x) = — = x'", then f'{x) = -njc""-' = -^
88. f(t) = f - 3, [2, 2.1]
fit) = 2t
Instantaneous rate of change:
(2, 1) => /'(2) = 2(2) = 4
(2.1, 1.41) => /'(2.1) = 4.2
Average rate of change:
/(2.1)-/(2) 1.41-1
84. True. If y = x/n = (I/tt) • .x, then rfy/otc = (l/7r)(l) = I/tt.
2.1 - 2
0.1
4.1
92.
s(t) = - 16/2 - 22r + 220
v(f) = -32r- 22
v(3) = -118 ft/sec
s(t) = - 16f2 - 22f + 220
= 112 (height after fallmg 108 ft)
- 16/2 - 22/ + 108 = 0
-2(/ - 2)(8/ + 27) = 0
/ = 2
v(2) = -32(2) - 22
= - 86 ft/ sec
■[»■!]
90. fix) = sin j:,
fix) = cos;c
Instantaneous rate of change:
(0, 0) ^ /'(O) = 1
73
77 2.
6' 2
/'
0.866
Average rate of change;
/(7r/6)-/(0)_ (1/2) - 0 ^ 3 ^
(77/6) - 0 (7r/6) - 0 TT
94. s(t) = -4.9/2 + v'(,/ + 5o
= -4.9/2 + 5o = 0 when / = 6.8.
5o = 4.9/2 = 4.9(6.8)2 = 226.6 m
342 Chapter 2 Differentiation
96.
2 4 6 8 10
Time (in minutes)
(The velocity has been converted to miles per hour)
98. This graph corresponds with Exercise 75.
(0,0)
2 4 6 8 10
Time (in minutes)
100. s{t) = --at^ + c and s'{t) = -at.
Average velocity:
5(fo + M) - 5(^0 - ^t) _ [-(l/2)a(fo + M)^ + c\- [-(l/2)a(fo - Af)^ + c)]
% + Af) - (ro - Af) 2Af
-{\/2)a%^ + 2foAf + (Af)^) + (l/2)a(fo^ - IfpAf + (Af)^)
dV
102. V = j^ — = 3j2
as
dV
When 5 = 4 cm, —- = A% cm^.
ds
2Ar
- 2afoAr
2Af
-ato
s '(tg) Instantaneous velocity at f = r,,
104. C = (gallons of fuel used)(cost per gallon)
15,000\, ^ 18,750
dC^ 18,750
dx x^
X
10
15
20
25
30
35
40
C
1875
1250
537.5
750
625
535.71
468.75
dC
dx
-187.5
-83.333
-46.875
-30
-20.833
-15.306
-11.719
The driver who gets 15 miles per gallon would benefit more from a 1 mile per gallon increase in fuel efficiency.
The rate of change is larger when x = 15.
106. ^=K(T- TJ
Section 2.2 Basic Differentiation Rules and Rates of Change 343
108. >> = -, ;c > 0
X
At (a, b), the equation of the tangent line is
y = — t(x -a) OT y
a a
The j:-intercept is (2a, 0).
2
a^ a
The y-intercept is I 0
1 1 (1
The area of the triangle is A = -bh = -(2a) -
2 I \a
110. y = x^
y' = 2x
(a) Tangent lines through (0, a):
y - a = 2x(x - 0)
X- — a = 2x~
±V— a = X
The pwints of tangency are (± ^--a. - a). At ( V-a, - a) the slope is y '( -J— a) = 2 v— a. At i
>''(-y^^ = -2^-0.
Tangent lines: jy + a = 2V— a(A- — -J— a) and _y + a = — 2V— a(.v + v — a)
' V=-2, '
-a. - a) the slope is
-ax + a
y = 2^ —ax + a y
Restriction: a must be negative,
(b) Tangent lines through (a, 0):
y - 0 = 2x(a: - a)
7? = 2x^ — 2mx
0 = jr — 2ajc = x(x — 2a)
The points of tangency are (0, 0) and (2a, 4a-). At (0, 0) the slope is .v '(0) = 0. At (2a, ^<^) the slope is >■ '(2a) = 4<3.
Tangent lines: v - 0 = 0(.r - 0) and y - 4a- = 4a(.r - 2a)
>> = 0 V = 4a.r — 4fl-
Restriction: None, a can be any real number.
112. /i(j:) = Isinjcl is differentiable for all x i= mr, n an integer.
/2(jc) = sin|x| is differentiable for all .r i= 0.
You can verify this by graphing/, and/j and observing the locations of the sharp turns.
344 Chapter 2 Differentiation
Section 2.3 The Product and Quotient Rules and Higher-Order Derivatives
2. f{x) = (6x + 5)(x5 - 2) 4. g{s) = JKa - 5^) = s''\A - P)
fix) = (6. + 5)(3.^) + (^ - 2)(6) ^,^^^ ^ ^^^^^_^^ ^ ^^ _ ^,^1^_,,, ^ _^^3,, ^ l-_^^
= 18x3 + 15a:2 + 6x3-12
= 24x3 + 15^2 _ 12
6. g(x) = Vx sin X
g '(x) = Vx cos X + sin x( — -7= J = Vx cos x +
10. h{s)
4- 552
/s- 1
VW
25'/2
-»<"=g^
1
^ sinx
2Vx
,, . (2f - 7)(2f) - (r2 + 2)(2)
. '^'^- {2t-ir-
12.f(t) = y
2f _ 14; _ 4
(2f - If
,. t\- sm t) - cos t{Zt^)
r sin f + 3 cos r
(V^ - 1)(1) - .(|.-'/2J .w (^3)2
V^ - 1 — — v«
2^" V;-2
[Js-\f 2(7^ - if
14. /(x) = (x2 - 2x + l)(x3 - 1) 16. fix) = J^
^-(,) = (,. _ 2, + i)(3x^) + (x3 - l)(2x - 2) ^_^ - 1)(1) - (. + 1)(1)
= 3x2(x - 1)2 + 2(x - l)2(x2 + X + 1) ^^'''
= (x- 1)2(5x2 + 2X + 2)
/'(I) = 0
(x-
1)^
X — 1 — X -
- 1
(X - 1)2
2
{X - 1)2
2
= -2
^'(2)= (2-1)
18. /(x)=^
„, , (x)(cosx) - (sinx)(l)
/w-
X cos X - sin X
r2
/7r\ (7r/6)(V3/2) - (1/2)
■^U/ TJ-2/36
37377- 18
772
3(7377-6)
772
Function Rewrite
DenVarive
Simplifv
5x2-3 5 ,
20. y = ^ y- ^x^-
3
4
, 10
, 5x
Section 2.3 The Product and Quotient Rules and Higher-Order Derivatives 345
Function
Rewrite
22- y = L^
, = |.-
24..= ^^^-^
3 2 5
26. f{x) - ^ _ J
(.x= - mx^ +
3) - (x3 + 3x + 2)(2x)
J w -
U^ - 1)^
.t* - 6x2 _ 4^
-3
{x- - D-
30. fix) = ^(v^ + 3) = x>/3(x'/2 + 3)
* _1/-|\ . / !/'> . n\l _„ — 2/3
/'W = x^'H-^x'^^^j + U>/2 + 3)1 jx
= 7X-'/6 + x-2/3
6
' +^
6x1/6 ^2/3
Alternate solution:
/(X)= 3/^(v/^ + 3)
= x5/6 + 3x1/3
/'(x) = fx-'/6 + X-2/3
5 +^
Derivative
= --r-3
Simplify
y =
5x3
6x
y =7^
28. /W = -^
fix) = x^
= 2x3
x+ ij U+ ij
(x + 1) - (x - 1)1 [x - 1
ix + \r
:" + X - 2]
H
X + 1
(4.r3)
2x^
(.
32. /i(x) = (x^ - 1)= = x^ - 2x2 + 1
/i'{x) = 4x3 - 4x = 4x(x2 - 1)
(^U6 ^2/3
34. g{x)=xi- ^
= 2x -
g'W
X X + 1 / X + 1
(x + l)2x - x^d) _ 2(x2 + 1-c + 1) - x2 - 2x _ x2 + It + 2
U + 1)2
U + 1)^
(X + 1)2
36. /(x) = (x2 - x)(x2 + l)(x2 + X + 1)
fix) = [Ix - 1)(X- + l)(x2 + X + 1) + (x2 - X)(2x)(x2 + X + l) + (x2 - x)(x2 + l)(2x + l)
= (2x - l)(x'* + .v3 + 2x2 + X + 1) + (x2 - x)(2r3 + 2x2 + 2x) + (x2 - x)(2t3 + x2 + 2v + 1)
= 2x5 + x'' + 3x3 + X - 1 + 2x-5 - 2x2 + 2x^ - x'* + .t3 - x2 - X
= 6x5 + 4x3 - 3_^2 _ 1
38. /(x) = £1 ^■
fix) =
(c2+.r2)(-2v)-(c2-x2)(2v)
ic~ + X2)2
40. fie) = (e+ i)cos0
fie) = ie+ i)(-sin e) + (cos e)(i)
= COS e - (e + 1) sin e
— 4xc2
(C^ + x2)2
346 Chapter 2 Differentiation
42. fix) =
f'(x)
X COS X - sin ;i:
44. y = X + cotx
y' = 1 — csc^jc = -cot^;c
46. /z(j') = 10 CSC j'
h '(s) = — J + 10 CSC 5 cot ;
48. y
sec a:
, X sec X tan j: — sec x
y = 72
sec.t(xtanx — 1)
50. y = x sin x + cos x
y' = x cos ;c + sin ;c — sin X = jt cos x
52. /(j:) = sin x cos x
/'W = sin;c(-sinj:) + cos;c(cosx)
= cos 2x
54. h{e) = Sdstcd + dime
h'{e) = SOsec etan 0 + 5 sec e + Ssec^ 9 + tan e
58. /(e)
/'(e)
sin 6
1 - cos I
1
cos d — \
cos 0 - 1 ~ (1 - cos 9)2
(form of answer may vary)
56. fix) = (^VfT^)^^' "^ ^ + ^^
;f5 _|_ 2;i^ + 2;c2 - 2
f'(x) = 2 1 2 j^ \\2 (form of answer may vary)
60. f{x) = tan ;c cot x = 1
fix) = 0
/'(I) = 0
62. fix) = sin x(sin x + cos x)
fix) = sin j:(cos jt — sin ;c) + (sin x + cos x)cos x
= sin j: cos x — sin^ ;c + sin x cos x + cos^ x
= sin 2x + cos 2x
lir\ . 77 . •"■ 1
64. (a) fix) = ix- \)ix-^ - 2), (0, 2)
fix) = ix- 1)(2a:) + ix^ - 2)(1) = Zx^ - 2x - 2
/'(O) = -2 = slope at (0,2).
Tangent line: y - 2 = -2x => y = -2x + 2
(b)
66.(a)/(.) = ^,(2,|
. ^ ix + 1)(1) - (x - 1)(1) ^ 2
■^ ^ ^ (;c + 1)2 ix + 1)2
/'(2) = I = slope at (2, | ,.
12 2 1
Tangent line: y - - = -ix - 2) =!> y = -x - -
(b)
y
___
/
Section 2.3 The Product and Quotient Rules and Higher-Order Derivatives 347
68. (a) fix) = sec;c, \^-,2
fix) = sec jt tan jf
/||j = 2y3 = slopeat^|,2J.
Tangent line:
673x -3y + 6- 1^-n = 0
70. fix)
x^+ 1
.,, , _ U^ + l)(2x) - (x^)(2x) Ix
fix) = 0 when x = 0.
Horizontal tangent is at (0, 0).
(.V- + 1)2
(b)
V.
/
^^
/ ^
-.-. ,-,/ X xizosx - 3) - (sinj: - 3a:)(1) xcosx — sinx
72. / (x) = ^3 = —^
,, , x(cosx + 2) — (sinx + 2x)(l) xcosx — sinx
g U) = ; = 1
, , sin X + 2x sin X — 3x 4- 5x ,, .
gix) = = = fix) + 5
X X
/ and g differ by a constant.
_. ., . cosx
74. fix) = = X " cos X
•^ X"
/'(x) = — x~" sinx — MX"""' cosx
= — x"''"H^sinx + ncosx)
X sin X + « cos X
x" +
76. V = 7rr2/i= -rrit + 2)(^Jt
= |(r3/2 + 2ri/2)^
V"(f) = ^tI:^''^' + t"'''- W = -7-775-77 cubic inches/sec
2\2
4ri
When/! = 1: fix) = -
X sin X -(- cos X
When n = 2: /'(x) = -
When /I = 3: /'(x) =
When « = 4: fix) =
X sin X + 2 cos x
x3
X sin X -^ 3 cos x
X sin X + 4 cos x
78. P = -
^ , ^,/ > X sin X + n cos x
For general njix) = 1^7:^ .
80. fix) = sec X
^(x) = CSC X, [0, 27r)
fix) = g'ix)
1 sinx
sec X tan X , cosx cosx
sec X tan X = — esc x cot x => \ — = — 1
sin^x
cos^x
317 Itt
CSC X cot X
= - 1 =* tan^ X = - 1 =* tan X = - 1
1 cos X
sin X sm X
-1
348
Chapter 2 Differentiation
82. (a)n(f) = -9.6643r2 + 90.7414f + 77.5029
v(f) = -276.4643r2 + 2987.6929/ + 1809.9714
vW ^ -276.46f- + 2987.69f + 1809.97
^ -* «(f) -9.66/2 + 90.74f + 77.50
A represents the average retail value (in millions of
dollars) per 1000 motor homes.
40A6{x^ - 2.09;c + 17.83)
(c) A (t) ,^2 _ ^ 3^^ _ g Q2)2
ix'
,/ s x-^ + 2j: — 1
86. fix) = = X
fix) = 1+-,
/'« = -^
2-i
84. fix) =x\^
x'-
fix) = 1
64
fix)
192
88. /W = secx
fix) = secjctanx
/"W = sec jc(sec^ jc) + tan A;(sec j: tan ;c)
= sec ;c(sec^ x + tan^ x)
90. fix) = 2 - 2x-'
/'"(x) = 2x-2 = 4
92. /<'»'W = 2;c + 1
/'='W = 2
/fe)W = 0
94. The graph of a differentiable func-
tion/such that/ > 0 and/' < 0
for all real numbers x would in
general look like the graph below.
96. fix) = 4 - hix)
fix) = -h'ix)
/'(2) = -;!'(2) = -4
98. fix) = g(;c)/r(jc)
/'(;t) = gix)h'ix) + hix)g'ix)
/'(2) = gi2)h'i2) + hiDg'H)
= (3)(4) + (-l)(-2)
= 14
100.
It appears that/ is quadratic; so/'
would be linear and /"would be
constant.
102. sit) = -8.25/2 + 66/
v(/) = - 16.50/ + 66
ait) = - 16.50
/(sec)
0
1
2
3
4
sit) (ft)
0
57.75
99
123.75
132
v(/) = 5 '(/) (ft/sec)
66
49.5
33
16.5
0
a(/) = v'it) (ft/sec2)
-16.5
-16.5
-16.5
-16.5
-16.5
Average velocity on:
[0, 1] is ^^j^= 57.75.
[1,2] is 2^-15^.4,25.
[2,3]isi^Ml_99^24^3
M32- 123.75 ^g^3_
4-3
Section 2.3 The Product and Quotient Rules and Higher-Order Derivatives 349
104. (a) f(x) = x"
f (x) = n(n - l)(n - 2) ■ ■ • (2)(1) = n\
Note: n\ = n(n - I) ■ ■ -3 • 2 • \ (read "n factorial.")
106. [xf{x)]' = xf'ix) +f(x)
[xfix)]" = xf"{x) +f'{x) +f'{x) = xf"{x) + 2f'{x)
[xfix)]'" = xf"'ix) +f"{x) + 2f"(x) = xf"'ix) + 3/"U)
In general, [xf{x)}"^ = xf^"Kx) + «/'"-"U)-
108. f(x) = sin x
4f) =
/'(;c) = cosx fU =^
f"ix) = -smx /"(jj^"^
(a) FlU) =r{a)ix - a) + f{a) = 0[j: - -1 + 1 = 1
(b) f(x) = -
("jfr^ =
/'"'W
(-l)"(n)(n- !)(» - 2) ■ ■ ■ (2)(1)
/'2W = ^/"(fl)U - ar +f'(a)(x - a) + f(a) = -(- 1)U - - + 1
1 l/ ^
2\ 2
(c) Po is a better approximation than P,.
(d) The accuracy worsens as you move farther away from x = a
(-!)"«!
(b)
(?■')
IT
/"'N
V
V
110. True, y is a fourth-degree
polynomial.
112. True
d"y
dx"
0 when n>A.
116. (a) (fg ' - f'g) '=fg" + f'g'- f'g '-f'g
= fg"-f"g True
(b) (fg)"={fg'+f'g)'
= fg" + f'g' + f'g' + f'g
= fg"+2f'g'+f"g
=^fg" + f"g False
114. True. If v(f) = c then
a(t) = v'(r) = 0.
350 Chapter 2 Differentiation
Section 2.4 The Chain Rule
y=figix))
4. y = 3 tan(T7x^)
gix)
U = X + I
y=fiu)
y = M-'/2
y = Jitanu
6. y = cos
3x
3x
2 2
8. y = {2^+ \Y
y' = 2(2^' + l)(6;c2) = nx\2>? + 1)
12. /(f) = (9f + 2)2/3
/^(.) . |(9. . 2)-./3(9) ^ ^^
y = cos M
10. y = 3(4 - ^2)5
y' = 15(4 - x'^)\-2x) = -3044 - x'^Y
1/2
14. g(x) = V5 -3x = (5 - 3;c)
.w4(5-3.)-(-3) = ^^^
16. g(x) = V;c2 -lx+ \ = J{x - 1)2 = |x - 1|
1, X > 1
^'(^) =
•1, JC < 1
18. f{x) = -3V2 -9x
fix) = -3(2 - 9x)V4
/'(.) = -|(2-9.)-3/^(-9)=^(^^
20. s{t)
1
;2 + 3f - 1
40 = (;2 + 3f - 1)-'
j'(0 = -l(f2 + 3f- \)-\it + i)
-(2f+3)
(f2 + 3f - 1)2
22. y =
(f + 3)3
y = -5{t + 3)-3
y' = \5{t + 3)-'' =
15
(f + 3)^
24. .w=vS ;' ,,
g(f) = (f2 - 2)-'/2 ,
g'it) = -^(/2 - 2)-3/2(2f)
(,2 _ 2)3/2
26. /(x) = x{-ix - 9)3
/'(x) = x[3(3x - 9)2(3)] + Ox - 9)3(1)
= (3x - 9)2[9x + 3a: - 9]
= H{x - 3)2(4x - 3)
28. V = 42V16 - z2
y' = |^'(|(16 - x2)-'/2(-2x) ) + x(16 - x2)i/2
-^
2yi6"
+ Wl6 - x^
-xQx^ - 32)
2Vl6 - x2
30. y =
Tzn
1,
(x4+ 4)1/2(1) - x^(x^+ 4)-'/2(4x3)
x*+ 4
x* + 4 - 2x'' 4 - jc"
(jc" + 4)3/2 (^4 + 4)3/2
32. hit) =
,2 \2
?3 + 2
h'it) = 2( 3
f2 \/(f3 + 2)(2f) - f2(3f2)\ ^ 2g2(4f - f^) ^ 2r3(4 - ;3)
f3 + 2/1 (f3 + 2)2 / (f3 + 2)3 (r3 + 2)3
Section 2.4 The Chain Rule 351
«-« = (f^)'
Y ^ J3x^ - 2\V(2x + 3)(6x) - {3x^ - 2)i2)\
8 W = ^I^TTTT) I ?^7TTu j
2;c + 3
(2;t + 3)2
3(3;c2 - 2)2(6.t' + 18a: + 4) ^ 6(3jc2 - 2)-(3x2 + 9a: + 2)
(2a: + 3)'' (2a: + 3)''
36. y
^/
2x
x+ 1
1
^ 72^(X + 1)3/2
y ' has no zeros.
38. f(x) = 7^(2 - a:)2
(;c - 2)(5;c - 2)
/V) =
2v^
The zeros of/' correspond to the points on the graph of
/ where the tangent lines are horizontal.
y
W
40, y= it-- 9)Vr + 2
, 5t^ + 8r - 9
y =
2jt + 2
The zero of y ' corresponds to the point on the graph of y
where the tangent line is horizontal.
//
K
/ j
42. g{x) = V.v - 1 + V.r + 1
g'(x)
1 + »
2Vx - 1 2jx + 1
g ' has no zeros.
44. y = a:^ tan -
X
dy ^ \
-r = 2a: tan -
,1
sec--
X X
The zeros of y ' correspond to the points on the graph of
y where the tangent lines are horizontal.
W
46. (a) y = sin 3a:
y ' = 3 cos 3a:
y'(0) = 3
3 cycles in [0, 2Tr]
(b) y = sin!
;)cos(
y'(0) = ^
HaIfcyclein[0. 27r]
The slope of sin ax at the origin is a.
352 Chapter 2 Differentiation
48. V = sin TTx
dy
-f- = TT COS TT
dx
50. h{x) = sccix'^)
h '(x) = 2x secU^) tan{x^)
52. y = cos (1 - 2a;)2 = cos ((1 - 2x)^)
y' = -sin (1 - 2xY(2(\ - 2x)(-2)) = 4(1 - 2x) sin(l - 2xY
54. g{e) = secf ^ej tm[^9
g'ie) = sec(|ej sec2(|ej| + tan(|ej sec(|0J tan(^0j^
4secge[sec^|eUtan^ie
-, / X cosy
56. g(v) = = cos V • sm V
CSC V
g '(v) = COS v(cos v) + sin v(— sin v) = cos^ v — sin^ v = cos 2v
58. y = 2 tan^ x
y ' = 6 tan^ x • sec^ x
60. g(t) = 5 cos^ irr = 5 (cos Trrp
g'(f) = 10 cos 7rf(— sin Trt)i'7r)
= — 10iT(sin -n-t)(cos TTf) = — Sirsinlirf
62. h(t) = 2cot2(7r/ + 2)
/;'(«) = 4cot(7rr + 2)(-csc2(T7r + 2)(7r))
= -ATTCOtiTTt + 2) csc2(Trf + 2)
64. y = 3x - 5 cos(7rj:)^
= 3;i: - 5 cos(7r^;t2)
£ = 3 + 5sin(ir2x2)(2i72x)
= 3 + 10tt2;c sin(i7:t)2
66. .V = sin;c'/3 + (sinx)'/^
y' = cos ;c'/'( -x~^/^ + —(sin a:)~^/' cos x
cosx'/^ COSJf 1
;c2/3 "^ (sin;c)2/3j
68. >- = (3x3 + 4^)i/5_ (2, 2)
>'' = |(3x^ + 4x)-''/5(9x2 + 4)
9x- + 4
5(3x3 + 4jj)4/5
y(2) = 2
70. /W = (^^ = (.- 3x)- (4,^
fix) = -2(x2 - 3x)-3(2x - 3)
^'(^) = -32
-2(2x - 3)
(x2 - 3x)3
72. /(x)=^^, (2,3)
fix)
2x- 3'
(2x - 3)(1) - (x + 1)(2)
(2x - 3)2
(2x - 3)2
/'(2) = -5
TA 1 ^ / h- 2
74. V = — I- Vcos X, — , —
X \2 IT.
1
-f2 2 Vcos X
y'i'jT/2) is undefined.
Section 2.4 The Chain Rule 353
76. (a) fix) = |Wx2 + 5, (2, 2)
fix) = ^;c
(x2 + 5)-'/2(2jc)
+ ^U^ + 5)'/2
+ ^Jx' + 5
3V^?T5 3
/'(2)^^4(3) = ^
Tangent line:
>■ - 2 = y (x - 2) => 13x - Q)' - 8 = 0
(b)
^
IZ
78. (a) f(x) = tan^x, i^, 1
/'(x) = 2 tan X sec^ x
/'(l) = 2(1)(2) = 4
Tangent line:
(b)
y-i=4U--
A
4x - y + (1 - tt) = 0
80. /W = (x-2)-'
/'W = -(x-2)-2 =
fix) = 2U - 2)-3
1
U - 2)^
2
(x - 2)3
82. /(x) = sec^ TTX
fix) — 2 sec 7rx(ir sec ttx tan ttx)
= 2ir sec^ ttx tan i7x
/"(x) = 277 sec^ 7rx(sec^ 7rx)(7T) + 27rtan 7rx(2Trsec' ttx tan ttx)
= 2ir^ sec'' ttx + 47r^ sec^ ttx tan^ ttx
= 2-77^ sec^ TTx(sec- ttx + 2 tan^ vx)
= Itr- sec'^ '77x(3 sec' t7x — 2)
84.
86.
/is decreasing on (— oo, — 1) so/' must be negative
there. /is increasing on (1, oo) so/' must be postive
there.
The zeros of/' correspond to the points where the graph
of/has horizontal tangents.
88. g(x)=/(x2)
g'(x) =/'(x-)(2x) => g'(.v) = 2X-/V-)
354 Chapter 2 Differentiation
90. (a) g{x) = sin2;c + cos^;^ = 1 ^> g'W = 0
g '(x) = 2 sin j: cos x + 2 cos x(— sin x) = 0
(b) tan^x + 1 = sec^.n:
g{x) + l^f(x)
Taking derivatives of both sides,
Equivalently,/'(x) = 2 sec x • sec x • tan x and
g '(x) = 2 tan X • sec* x, which are the same.
92. y = \ cos 12f - 5 sin 12r
V = >-' = |[- 12 sin 12f] - \[U cos 12r]
= -4 sin lit - 3 cos 12f
When t = 77/8, y = 0.25 feet and v = 4 feet per second.
94. >• = A cos wt
3.5
(a) Amplitude: A = -r- = 1.75
y = 1 .75 cos wr
r. ■ J in 277 TT
Penod: 10 => w = — — = —
V = 1.75 cos — r
(b) V = >>'= 1.75
77 . irf
= -0.35 77 sin
77f
96. (a) Using a graphing utility, or by trial and error, you
obtain a model of the form
(TTt
T{t) = 64.18 - 22.15 sini — + 1
(b)
(c) rW = -22.15 cos(y+l)(|
TTt
= -11.60 cost— + 1
(d) The temperature changes most rapidly when r = 4.1
(April) and t = 10.1 (October). The temf>erature
changes most slowly (T'(t) = 0) when r = 1.1
(January) and r = 7.1 (July).
98. (a) g{x) =f{x)-2^g '{x) = f'(x)
(b) h{x) = lf{x)=^h\x) = 2f\x)
(c) Ax) =f{-3x)=>r'{x) =f'i-3x){-3) = -3/'(-3x)
Hence, you need to know/'(-3j:).
r'(0) = -3/'(0) = (-3)(-|) = 1
r'(-l) = -3/'(3) = (-3)(-4)= 12
(d) six) = f(x + 2) ^ s'ix) = fix + 2)
Hence, you need to know/'(A: + 2).
s'{-2)=fX0)
X
-2
-1
0
1
2
3
/'W
4
2
3
1
3
-1
-2
-4
g'ix)
4
2
f
1
3
-1
-2
-4
h'{x)
8
4
3
~3
-2
-4
-8
r\x)
12
1
s'ix)
1
3
-1
-2
-4
-3. etc.
Section 2.4 The Chain Rule 355
100. fix +p)= fix) for all x.
(a) Yes,/'(j: + p) = fix), which shows that/' is
periodic as well.
(b) Yes, let gix) = filx), so g'ix) = 2/'(2x).
Since/' is periodic, so is g'.
102. \ffi-x) = -/(x), then
£[/(-^)] = f [-/W]
/'(-x)(-l) = -/'(x)
/'(-x)=/'(x).
Thus,/'(x) is even.
104.
|[|«|] = j^[J^] = |m-'/^(2««0 = ^ = „'^.„ ^ 0
106. /{x) = |x2 - 4|
108. fix) = |sinx|
/'(x) = cos x( 7-; — r ],x¥'klT
' sinx '
110. (a) fix) = sec(2x)
/'(x) = 2(sec2x)(tan2x)
fix) = 2[2(sec 2x)(tan 2x)] tan 2x + 2(sec 2x)(sec2 2x)(2)
= 4[(sec 2x)(tan- 2x) + sec-* 2x]
(b)
4f)=K?
il
2sec|f tanf =4V3
/"( Ij = 4[2(3) + 2^] = 56
P,(x) = 4V3(x - -f j + 2
P^Cx) = |(56)(x - 1)' + 4^3 (x - f) + 2
= 281 X - -^ r + 4V^( X - ^ I + 2
TT\2
6y
(c) P2 is a better approximation than P^
112. False. If/(x) = sin^ It, then/'(x) = 2(sin lx)(2 cos Ix).
(d) The accuracy worsens as you move away
from X = ir/6.
114. False. First apply the Product Rule.
356 Chapter 2 Dijferentiation
Section 2.5 Implicit Differentiations
2. x^-f=\e
2x - lyy' = 0
, X
6. x'y + y^x = - 2
x^' + 2xy + y^ + lyxy' = 0
{x^ + 2xy)>''= -{y^ + Iry)
y
-y(y + 2t)
;c(;c + 2y)
jc' + / = 8
3;c2 + 3y2y' = 0
(xy)'/2 - X + 2^ = 0
\{xy)-'l\xy' + y) - 1 +2>''=0
-^y' + -^
1 + 2)-' = 0
2v^ ijxy
xy' + y - ijxy + 4v^>'' = 0
2V^-y
4vxy + X
10. 2 sin X cos y = 1
2[sinx{-siny)y' + cosy(cosx)] = 0
cos X cos y
y = ^ :
sin X sm y
= cot X cot y
12. (sin TTx + cos Try)^ = 2
2(sin TTX + C0S7ry)[Tr cos ttx — 7r(siniTy)y'] = 0
77 cos TTX — TT^sin '77y)y ' = 0
COS TTX
y =
Sin Try
14. cot y = X — y
(— csc^y)y' = 1 — y'
1
y' =
1 — csc^y
1
-cot-y
= -tan-^y
16. X = sec-
y' 1 1
1 = —=^ sec- tan -
r y y
sec(l/y) tan(l/y)
-y2cos(^)cot(^)
18. (a) (x2 - 4x + 4) + (y2 + 6y + 9) = -9 + 4 + 9
(x - 2)2 + (y + 3)2 = 4 (Circle)
(y + 3)2 = 4 - (x - 2)2
(b)
= -3±V4 - (x - 2)2
(c) Explicitly:
^ = ±^[4-(x-2)2]-i/2(-2)(x-2)
ax 2
^(x
-2)
(74-
(;c - 2)2
-(^
-2)
±V4-
(;c - 2)2
-(;c-2)
-3 ± 74 - (x - 2)2 + 3
-(;c-2)
y + 3
(d) Implicitly:
2x + 2yy'- 4 + 6y'= 0
(2y + 6)y' = -2(x - 2)
, -{x-D
Section 2.5 Implicit Differentiation 357
20. (a) 9y'^
x^ + 9
(b)
y^ = ^+l =
x^ + 9
±Jx^ + 9
.^p ,., ^y 4^'^'r^''^'^^ ±x ±x X
(0 ExpLctly: - = = j^^=== = ^(^ = ^
(d) Implicitly: Qy^ - x^ = 9
ISyy' - 2x = 0
18>7'=2x
2x a:
ISy 9y
22.
/ = 0
2i - 3y^' = 0
3/
y =
At (1,1): y' = -.
24. (x + yy = }^ + f
x^ + 3;c^ + 3xf + f = x^ + y^
3x^ + 3ry2 = 0
x^y + xy^ = 0
x^' + 2xy + 2xyy' + y^ = 0
{x- + 2xy)y'= -(y^ + 2xy)
y(y + It)
"jc(.t + 2y)
At(-1, 1): y' = -1.
26.
X3 +
4ry + 1
3j;^ + 3y^y ' = ^xy ' + 4v
(3>'^ - 4x)y ' = 4>' — 3x2
, 4y - 3x2
At (2, !),>.' =
(3y2 -- 4x)
4-12 8
28. X cos y = 1
x[— y'siny] + cosy = 0
y -
cos y
X sin V
1 cot V
= — cot V = -
X ' X
*"^'f>>' = 5^-
30. (4 - x)y2 = x3
(4-x)(2)yO+r(-l) = 3x2
3X2 + y2
y =
2y(4 - x)
At (2, 2): y' = 2.
32. x' + y3 - drv = 0
3x2 + 3^^,' - 6,rv.' - 6y = 0
y'(3>-2 - dv) = 6y
, 6v
3.x2
3.t2
3^•2 - dr
M 8\ , ^ (16/3) - (16/9) ^ 32
'^^ \y ir ■ (64/9) - (8/3) 40
358 Chapter 2 Differentiation
34. cos y = X
— sin y • y ' = 1
•1
y ' = -: — , 0 < V < TT
sin 3;
sin^y + cos^y = 1
sin-y = 1 — cos^y
siny = Vl — cos-^y = Vl - x^
-1
^ yn^'
36.
-1 < ;c < 1
xY - 2x = 3
2x^' + 2xr - 2 = 0
jc^yy ' + xy^ —1=0
1 -xy^
->
xy
2xyy' + x2(y'P + x^" + 2xyy' + / = 0
4xKy' + x^iy^ + x^" + / = 0
4-4xy^- , (1 - xy^)^ , 2, - , 2 n
=^ H 7-7 1- x'Vy + y^ = 0
X xy^
4xy^ - 4xY + 1 - 2xy2 + ^.ly^ + ;t4y3y" + ^2^4 ^ q
x'^y^y" = 2xy — 2xy~ — 1
„ 2xY -2xf-\
38. \ — xy = X — y
y — xy = x — 1
X- 1
y' = 0
/'=0 ■
A'
40. / = 4^
2yy' = 4
, 2
y"=-2y-Y =
r-2]
LrJ
2 _ -4
y y
, X — 1
42. >^ = JTT
2>y'
(x^ + 1)(1) - U - l)(2x)
{x^ + 1)2
X2+ 1
- 2x2 + 2x
(;c2
+ 1)^
1 + 2x
-x2
2y{x2 + 1)2
1+4-4
1
"^''^•5 }"' [(275)/5](4 + 1)2 1075-
V5
1
Tangent line: > ~ ~^ = /^U - 2)
lOVSy - 10 = X - 2
X - lOySy + 8 = 0
Section 2.5 Implicit Differentiation 359
44. a:^ + y^ = 9
At (0, 3):
Tangent line: ^^ = 3
Normal line: x = 0.
At (2, VS):
_2
Tangent line: v - Vs = —i={x - 2) => 2a: + VSy -9 = 0
Normal line: y - V5 = -^{x - 2) ==> VSJc - 2y = 0.
(0.31
1
f A
V
y
/--
>t'
>i
J^
46. >»* = 4x
2xy'=4 , _
y' = -= 1 at (1,2)
Equation of normal at (1, 2) is y - 2 = -\{x — 1), v = 3 - x. The centers of the circles must be on the normal and at
a distance of 4 units from (1,2). Therefore,
[x - 1)2 + [(3 -x)- If = 16
2(a: - \)- = 16
x= \ ± ijl.
Centers of the circles: (l + 2^2, 2 - 2V2) and (l - 2^2, 2 + 275)
Equations: [x - \ - ijl)- + (v - 2 + ijl]- = 16
(x - 1 + ijlf + (.V - 2 - 272)- =16 -
48. 4x2 + y - 8x + 4.^, + 4 = 0
8x + lyy' - 8 + 4y'= 0
8x 4 - 4x
" 2>' + 4 .V + 2
Horizontal tangents occur when x = 1 :
4(1)2 + y2 - 8(1) + 4^ + 4 = 0
y2 + 4y = y(>' + 4) = 0 => y = 0, -4
Horizontal tangents: (1, 0), (1, -4).
Vertical tangents occur when y = —2:
4a- + (-2)2 - 8x + 4(-2) + 4 = 0
4x2 - 8.T = 4x(x - 2) = 0 => X = 0, 2
Vertical tangents: (0, -2), (2, -2).
360 Chapter 2 Differentiation
50. Find the points of intersection by letting y^ = x^ in the equation 'bp- + 3>'-^ = 5.
Ix'^ + lx^ = 5 and 3x5 + Ix^ - 5 = 0
Intersect whenx = 1.
Points of intersection: (1, ±
1)
/ = ^:
2x2 + 3^2 -.
= 5:
2yy' = 3a:2
Ax + 6yy ' =
= 0
2x
'^-yy
At (1, 1), the slopes are:
r-\
2
y =-3.
At (1, - 1), the slopes are:
3
y =-2
, 2
^ =3-
Tangents are perpendicular.
[21^+3/ =5]
/"
~>((1. 1)
v^
[>=;■
52. Rewriting each equation and differentiating,
jc^ = 3(y - 1) xOy - 29) = 3
^ ,
. = y + l
>» = jj"'
-K!-'
3' =
1
|jr(3>-29) = 3|,5 [x_3=3>.-
A
A
J
^
f
For each value of jt, the derivatives are negative reciprocals of each other. Thus, the tangent lines are orthogonal at both points
of intersection.
54.
X2 +y2= C^
y = Kx
2x + 2yy' = 0
y' = K
• X
'^~y
c
7K'
k
X
At the point of intersection {x, y) the product of the
slopes is {—xly)(fC) = (—x/Kx){K) = — 1. The curves are orthogonal.
K = -] \^-^ c =
56. ^2 - 3xy2 + y = 10
(a) 2;c - 3/ - 6xyy' + 3y^' = 0
i-6xy + 3/)y'= 3y2 - 2;c
y=3/-2.
^ 3y^ - 6xy
*>-f-*'f-'^vf = °
(2x - 3/)^ = {6xy - 3y^^
58. (a) 4 sin x cos >> = 1
4 sin x(— sin y)y ' + 4 cos x cos y = 0
cos X cos y
>" = —■ ■■
sin X sin y
I \dy dx
(b) 4smjc(— siny)— + 4 cos x— cosy = 0
at at
dx
.-.^dy
cos JT cos y-r = sinjcsiny-,
dt dt
Section 2.6 Related Rates
361
60. Given an implicit equation, first differentiate both sides
with respect to x. Collect all terms involving y' on the left,
and all other terms to the right. Factor out v ' on the left
side. Finally, divide both sides by the left-hand factor that
does not contain y '.
62.
64.
>/x + Vv = v/c
' +-^^=0
2Vx iVy'^
di
dx
V-y
Tangent line at (xq, yg):
yo
^U - Xg)
77jn
Use starting point B.
.x-intercept: [xg + v^Vyo> O)
y-intercept: (o, Vq + V^v5o)
Sum of intercepts:
Uo + V^VS^) + (.Vo + v^v^) = -^o + 2V^V>\J + Jo = (^ + n/v^)" = (Vc)" = c.
Section 2.6 Related Rates
2. y = 2(x2 - 3x)
Tt^^^'-^^t
dx
1 dy
dt 4x - 6 dt
(a) When A; = 3and^ = 2,^ = [4(3) - 6](2) = 12
(b)When.= ,andf=5,f = ^(5) = -f
4.
.r^ + / = 25
2x
dx
dt
■ly
di
dt
di
dt
dx
dt
x^x
yldt
y\dy_
' xldt
(a) When x = 3, y = 4, and dx/dt = 8.
dt 4
(b) When .t = 4, y = 3. and dy/dt = -2.
dx
dt
r(-2)
362 Chapter 2 Dijferentiation
6. y
1 +x2
dx
dt
= 2
-2x
dx
dy_
dt L(l+x2)2jdr
(a) Whenx = -2,
dy -2(-2)(2) 8 ,
■ = :^ = TT^ cm/sec,
dt 25 25 '
(b) When x = 0,
— = 0 cm/ sec.
ar
(c) When x = 2,
dy _ -2(2)(2) _ -8
dr
25
25
cm/sec.
8. y = sinj:
dx
, = cos X—r
dt dt
(a) When x = 7r/6,
= ( cos — 1(2) = >/3 cm/sec.
(b) When x = 7r/4,
— = (cos— 1(2) = V^ cm/sec.
(c) When x = 7r/3,
-j- = (cos-r-j(2) = 1 cm/ sec.
dx . dy .
10. (a) — negative ==> — negative
dt dt
„ dy . . dx
(b) — positive => — positive
dt dt
12. Answers will vary. See page 145.
14. D = Jx^ + /
dt
j£^
— — = - (x^ + sin^ x) '/^{2x + 2 sin X cos x)^
flf 2 dr
16. A = trr^
dx jc + sin j: cos x dx 2 + 2 sin a; cos x
dA , dr
— - = lirr—
dt dt
If dr/dt is constant, dA/dt is not constant.
— - depends on r and -r.
dr dr
v^
+
sin-
;c dr
18.
V
dr
dt
dV
dt
Vx2 +
= 2
= 4Trr2
sin"^x
dr
dr
dK
(a) When r = 6, — = 477(6)^(2) = 28817 in Vmin.
dt
dV
When r = 24, — = 4ir(24)2(2) = 460877 inVmin.
(b) If dr/dt is constant, dV/dt is proportional to r^.
20. K = ;c3
dr
dV
- = 3;c2 -
dr dr
(a) When ;c = 1
dV
dt
= 3(1)2(3) = 9 cmVsec.
(b) When x = 10,
dV
dr
= 3(10)2(3) = 900cmVsec.
Section 2.6 Related Rates 363
22. V = ^TTr% = ^TrrK3,r) = Trr^
dt
dV - ^dr
dt dt
24. v = -.r^;, = -,.— ;,3=__,3
/ r h 5 , \
I By similar triangles, - = — => r = —h.j
dt
(a) When r = 6,
dt
= 3-7r(6)2(2) = 21677 inVmin.
(b) When r = 24,
dV
dt
= 377-(24)2(2) = 345617- inVmin.
fify ^ ^^2 ^ ^ = / 144 W
</? ~ 144 dt^ dt~ \25TTh^jdt
dh \44 9
26. V = -foA(12) = 6bh = 6^2 (since b = h)
( ) ^= nh— — = — —
When /j = 1 and — - = 2, -- = -r77r(2) = - ft/min
fi!/ dt 12(1) 6
(b) If -r = o in/min = — - ft/min and h = 1 feet, then
af o 32
f = (l^^^) = !^'Vmin.
28. a:^ + y^ = 25
„ dx „ dy
2x— + 2y-Y = 0
dt dt
dx y dy O.lSy . dy
— - = —^•-f= ^ smce -r- = 0.15.
dt X dt X dt
When X = 2.5,
y = yi8J5,^ = -^41^ 0-15 = -0.26m/sec
dt 2.5
30. Let L be the length of the rope.
(a)
L2 = 144 + x^
^.dLdx
2L-— = 2x-r
dt dt
dx L dL AL . dL
— = — = smce -3- = -4ft/sec.
dt X dt X dt
WhenL= 13,
>;4fl/sec
X = JL^ - 144 = Vl69 - 144 = 5 ^
i!|iy
dx 4(13) 52 ^Q_
dt 5 5 - 10.4 ft/sec.
" i^ft''^
(b) If^= -4, andL = 13,
dt
Speed of the boat increases as it approaches the dock.
dL xdx
dt Ldt
= ^>-4)
^^--
AsL ^ 12 + ,
dL
dt
decreases, so
the speed decreases.
364 Chapter 2 Differentiation
32. x^ +y^ = 5^
^ dx „ ^ ds
2x— + 0 = 2s —
dt dt
dy
since -r = 0
dt
dx _ s ds
dt X dt
When s = \0,x= VlOO - 25 = V75 = 5 V3
^ = ^(240) = ^ = 16073 = 277.13 mph.
dt 5 V 3 V 3
. s^-
= 90^ + x2
x =
= 60
dx _
dt "
= 28
ds _
dt '
X dx
''~s"dt
Home
When X = 60,
s = V902 + 602 = 30VT3
d^
dt
60
30yi3
(28)
56
13
15.53 ft/sec.
36. (a)
20 _ y
6 y ~ X
20^-
- 20x = 6^
14y = 20a:
10
dx_
-5
rfr
£f> 10 (ic 10, ^, -50^,
^(y - x) _dy dx _-50 , ._ -50 35 _
15
ft/ sec
38. x{t) = - sin 77?, x^ + y2 = 1
277
(a) Period: — = 2 seconds
77
(b) When x = -y
VHI
2 4
Lowest point: 0
enx
3
= T0'^ =
v-
= 4 ^
3
10
3 .
= — sm 77/
1
^ sm 77/ = -
1
=^' = 6
dx _
dt
3
— 77 cos 77/
;c2 + / = 1
2x— + 2>'— = 0^— = 3-.
rf/ rf/ dt y dt
dy _ -3/10 3
15/4' 5
-977 -97577
'^"^■^ = 7Tp-I"'^°^'^
Speed =
2575 125
-9V577
125
= 0.5058 m/sec
Section 2.6 Related Rates 365
40.
1 = -1 J-
R A J R-i
dR^
dt
dR.
dt
1.5
R^' dt ~ /?,2 ■ dr "^ T?,^ ■ Jf
When /f , = 50 and /?2 = 75,
R = 30
42. rgtanO = v^
32r tan 0 = v^, r is a constant.
32rsec20— = 2v—
A dt
dv 16r
A
sec- 0 —
dt
Likewise, — = -rr- cos- 0
dt 16r
rff'
dt
1(50)2^'^"^ (75)2
= (30)-| 777^(1) + 7:;iT2(l-5)
0.6 ohms/sec.
44.
sin fl = —
X
^ = (-l)ft/sec
cos d\
de
dt
dd
dt
-10 dx
X- ' dt
10 dx
X- dt
{see e)
= ~'0f 25 ^ 10 1 ^ 2 ^ 2V2T
25- V25- - 10= 25572T 25^21 525
= 0.017 rad/sec
46. tane = -
— - = 30(2 tt) = 607Trad/min = 17 rad/sec
dt
sec^e
^\ ^ Udx
dtj 50\ dt
dx ,„ , Jdd
— = 50 sec- e\ —
dt \dt
(a) When 6 = 30°, ^ = ^^ ft/sec.
(b) When e = 60°, — = 200Tr ft/sec.
(c) When 6 = 70°,^ = 427.43Tr ft/sec.
48. sin 22°
0 = -— . ^ + i . ^
y- dt y dt
^ = - • ^ = (sin 22°)(240) = 89.9056 mi/hr
50. (a) dy/dt = 3(dx/dt) means that y changes three times as
fast as X changes.
(b) y changes slowly when v = 0 or .v = L. y changes
more rapidly when .v is near the middle of the
interval.
366 Chapter 2 Differentiation
52. L? = 144 + x^; acceleration of the boat
£x
dt^-
^- . ■ ■ ^r dL ^ dx
First denvative: 2L—- = 2x—-
dt dt
^dL dx
L-r- = X —
dt dt
, , . . d-L dL dL d^x dx dx
Second denvative: L -pr H — r ' ~r = ^<^tt '^ ~r ' ~T
dt^ dt dt dt^ dt dt
d^x
dt""
I'f-(f)"-(f
dx
dL
dL
d^L
When L = 13, j: = 5, -r = - 10.4, and —- = — 4 (see Exercise 30). Since —- is constant, -pr = 0.
dt
dt
dt
dt^
g = |[13(0) + (-4)2 -(-10.4)2]
= |[16 - 108.16] = |[-92.16] = - 18.432 ft/sec2
54. y(i) = -4.9r2 + 20
dy
dt
y(\) = -4.9 + 20 = 15.1
y\\) = -9.8
T, • M • , 20 y
By similar triangles, — = — r
X X — \2
20x - 240 = xy.
Whenjy = 15.1, 20;c - 240 = .x(15.l)
(20 - 15.1);c = 240
240
4.9'
20;t - 240 = ry
,„(& dy dx
^"^t^^t^^t
(0.0)1
Atr = 1,
dx
x
dy
dt
20 - y
dt
dx
240/4.9
I
dt 20 - 15.1
(-9.8) = -97.96 m/sec.
Review Exercises for Chapter 2 367
Review Exercises for Chapter 2
2. fix) =
x+ 1
X- 1
x + ^x + \ _ X + 1
,. fix + Ax) -fix) ,. ;c + Ax - 1 j: - 1
/'W = lim -^ — ^-^ = hm T
= lim
Ax->0
ix + Ax+ l)ix-\) - ix + Ax- Dix + 1)
AtU + Aj: - l)(jc - 1)
~ i^^o M^c + Ajc - l)(;c - 1)
lim
-2Ax
= lim
^"o Axix + Ax- 1)U - 1) ^"o (;c + Ajc - l)(x - 1) (jc - 1)-
4./W = -
■' Ai->o Ajc
6. /is differentiable for all jc t* —3.
lim
X + Ax X
Aj:-»0 Ax
2x: - (2x + 2Ax)
Aj:-»0 AjcU + Ajc)jC
-2Ax
Ax-^0 Ajc(x + Ajc)jc
r -2 -2
= lim 7 : — ^ — — TT
iijr->0 (X + Ax)x X~
, , _ Jx^ + 4x + 2, ifx < -2 '
»--^W-jj_4^_^2 if., >-2
(a) Nonremovable discontinuity at .V = — 2.
(b) Not differentiable at .r = - 2 because the function is
discontinuous there.
10. Using the limit defmtion, you obtain h '(«) = ^ - 4.t.
AtJC=-2,;z'(-2) = |-4(-2) = y.
-2
12. (a) Using the limit definition, fix) = -, -tt.
(jc + \y
At jc = 0./'(0) = -2. The tangent line is
>- - 2 = -2U - 0)
y = -Ix + 2
(b)
^
t^'
^
\
368 Chapter 2 Differentiation
14. /'(2) = lim
/U)-/(2)
2 X - 2
1
= lim
1
x+ I 3
= lim
2 JC - 2
3 -a:- 1
x->2 (x - 2)U + 1)3
-1
1
lim / , ,N-
x->2 (x + 1)3 9
18. y = - 12
y' = 0
20. gW=xi2
g'W = 12x"
16.
22. /(r)= -8^5
fit) = -40^
24. g(j) = 4s^ - 5s^
g'is) = I6s^ - \0s
30. g{a) = 4 cos a + 6
g '(a) = - 4 sin a
26. /(jc) = ;c'/2 - ;t-
1/2
28. h{x) = |x-2
/'W = |.-'/^ + 1.-/^ = ^
-4 _ -4
, , 5 sing
32. g(a) = — 2a
,. . 5cosa
g (a) = — 2
34. 5 = - 16f2 + So
First ball:
-16/2+ 100 = 0
Aoo 10 ^^ J u-
f = - / — -^ = — = 2.5 seconds to hit ground
V lo 4
Second ball: . ■ '
-16r- + 75 =0
f2 =
ns 573
2.165 seconds to hit ground
Since the second ball was released one second after the first ball, the first ball will hit the ground first. The second ball will hit
the ground 3.165 - 2.5 = 0.665 second later.
36. 5(f) = - 16;^ + 14,400 = 0
' 16/2 = 14400
r = 30 sec ■
Since 600 mph = g mi/sec, in 30 seconds the bomb will move horizontally (gjCSO) = 5 miles.
Review Exercises for Chapter 2 369
38.
(a) y
32 . ,,
32
— ^x
64
(b) v'= 1 -^x
0 if r = 0 or X
32'
Projectile strikes the ground when x = v^jlsl.
Projectile reaches its maximum height vXx = v^/M.
(one-half the distance)
32 / 32 \
(c) y = x ^.r' = x\\ ;.v = 0
when ,v = 0 and x = Xq-/32. Therefore, the range is
-v = Vq-/32. When the initial velocity is doubled the
range is
(2vo)^
32
32
or four times the initial range. From part (a), the
maximum height occurs when x = Vo^/64. The
maximum height is
y\
64
32 /vo'
64
Vo-\64
64 128 128
If the initial velocity is doubled, the maximum
height is
2 >
V
(2vo)
64
J 128 \128
(d) Vq = 70 ft/sec
iW-
Range: x = - ^^
153.125 ft
Maximum height: y = 7^ = ^:^ " 38.28 ft
or four times the original maximimi height.
40. (a) y = 0.14;c= - 4.43a + 58.4
(b) 320
(C) 2
(d) If.v = 65, V = 362 feet.
(e) As the speed increases, the stopping distance increases at an increasing rate.
370 Chapter 2 Differentiation
42. g{x) = (x3 - Zx){x + 2)
g\x) = (x3 - 3x)(l) + {x + 2)(3;c2 - 3)
= x^ - Ix + ix^ + 6x^ - 3x- 6
= 4.^ + ex'^ - 6x- 6
46./W = ^^l
f\x) =
^ {x- 1)2
50. fix) = 9(3^2 - 2x)-'
Z'U) = -9(3x2 _ 2x)-2(6x - 2)
54. y = 2jc — ;c2 fan ^^
y' = 2 — x^ sec^ x — 2xVwix
x -
- 1
{x-
- 1)(1) -
-(;c +
1)(1)
(x-
-IF
-2
18(1 - 3x)
(3a:2 - 2xY
44. /(f) = t^cost
fit) = t\-smt) + cosf(3f2)
= — f' sin f + 3?2 cos t
(x2 + 1)(6) - (6a: - 5)(2x)
(;c2 + 1)2
2(3 + 5x- ?,x^)
ix" + 1)2
fix)
sin x
52. y = ^
, _ (j:2)cosx — (sinx)(2x) _ x cos jc — 2 sin jr
xf>
X?
56. y
1 + sinx
1 — sin X
, _ (1 — sinx) cosx — (1 + sinx)(— cosx)
^ " (1 - sinx)2
2cosx
(1 — sinx)2
58. v(/) = 36 - t\0< t < 6
ait) = v'it) = -2r
v(4) = 36 - 16 = 20 m/sec
a(4) = - 8 m/sec
62. hit) = 4 sin f — 5 cos t
h 'it) = 4 cos t + 5 sin t
h"it) = -4 sin f + 5 cos t
60. fix) = Ux^'"'
fix) = 3X-3/4
-9 —9
^ W 4 ^ 4^7/4
64.
y
(10 — cosx)
xy + cos X = 10
xy ' + y — sin X = 0
xy ' = sin X — >>
xy' + y = (sin x — >>) + y = sin x
66. fix) = (x2 - l)'/3
fix) = |(x2 - 1)-V3(2x)
2x
3(x2 - 1)2/3
68. /(x) = (x2 + ^)'
Review Exercises for Chapter 2 371
70. h{e}
h'{9)
(1 - er
{i-ey-e[3(i- en-i)]
(1 - er
(1 - 8)^(1 - 6 + 36) ^ 26+ 1
(1 - ef (1 - 6Y
74. y = CSC 3x + cot 3x
y' — —3 CSC 3.t cot 3j: — 3 esc- 3x
= - 3 CSC 3x(cot 3x + CSC 3x)
72. y = I - COS 2j: + 2 cos^ x
y' = Is'mlx — 4 COS jc sin j:
= 2[2 sin j: cos x] - 4 sin x cos x
= 0
76. >■
sec^;
7 5
)> ' = sec^ a:(sec x tan x) — sec'' ar(sec j: tan x)
= sec^ X tan x(sec^ .t - 1)
= sec^ X tan^ x
78. f(x)
fix)
3x
Vx=+T
1,
3(x2 + l)'/2 - 3x|(x2 + l)-'/2(2x)
x2+ 1
3(x^ + 1) - 3x'
(X^ + 1)3/2
3
(x2 + 1)3/2
80. y
cos(x — 1)
X - 1
-(x - 1) sin(x - 1) - cos(x - 1)(1)
(x - ly-
1
U - 1)-'
[(x - 1) sin(x - 1) + cos(x - 1)]
82. fix) = [(x - 2)(x + 4)Y = (x2 + 2t - 8)2
fix) = 4(x3 + 3x- - 6x - 8)
= 4(x - 2)(x + l)(x + 4)
The zeros of/' correspond to the points on the graph of/ where the tangent line is horizontal.
w
84. ^(x) = x(x- + 1)1/2
2x2 + 1
g'ix)
VxH^
g' does not equal zero for any value of x. The graph of g
has no horizontal tangent lines. .
1
86. y = V3x(x + ly
3(x + 2)2(7x + 2)
y =
273x
y ' does not equal zero for any x in the domain.
The graph has no horizontal tangent lines.
f
372 Chapter 2 Differentiation
88. y = 2 csc^ (v^)
y' = ;pcsc^ Vxcot Vx
Jx
The zero of y ' corresponds to the
point on the graph of 3; where the
tangent line is horizontal.
T
90. y = j;-' + tanjc
y' = —x~'^ + sec^jr
y" = 2x~^ + 2 sec jc(sec x tan x)
= —; + 2 sec^ j: tan ;<:
96. h{x) = xjx^ - 1
2x^-1
x{2jP- - 3)
(;c2 - 1)3/2
92. y = sin^ x
y' = 2 sin X cos x
y" = 2 cos 2jc
sin 21
94. ,W = ^
;c2+ 1
2(-3a:^ + 5;c + 3)
(x?- + 1)2
(x2 + 1)3
98. V = J2^ = 72(32);; = 8v^
dh Jh
(a) When /i = 9, ^ = ^ ft/sec.
an 3
(b) When h = A,— = 2 ft/sec.
an
100. x2 + 9y2 - 4x + 3y = 0
2x + ISyy'- 4 + Sy' = 0
3(6y + l)y' = 4 - 2x
4 - 2x
y =
3(6y + 1)
102. ■f = x'-x^ + xy-y^
0 = x3 — x?y + xy — 2y^
0 = 3x2 _ j,2y ' _ 2xy + xy' + y — 4yy '
(x2 — X + 4y)y' = 3x2 _ 2xy + y
, _ 3x2 _ 2xy + y
y =
x^ — X + Ay
1©4. cos(x + y) = X
-(1 +y')sin(x + y) = \
— y'sin(x + y) = 1 + sin(x + y)
, _ 1 + sin(x + y)
sin(x + y)
= -csc(x + 1) - 1
106. x2 - y2 = 16
2x - 2yy' = 0
/ X
'^y
At (5, 3): y' = \
Tangent line: y - 3 = -(x - 5)
5x - 3y - 16 = 0
Normal line:y — 3 = - •r(x - 5)
3x + 5y - 30 = 0
108. Surface area = A = 6x2, j, length of edge.
dt
^ = 12x^ = 12(4.5)(5) = 270cmVsec
dt dt
Problem Solving for Chapter 2 373
110. tan e = ;c
—r = 3(27r) rad/min
2 Jd6\ dx
dx
dt
(tan2 e + l)(67r) = Gt^jc^ + 1)
When;c
1 <ic , /I ,\ IStt, , . ,^„ , ,,
= -, — = 0771 T + 1 = —z- km/ mil) = 450irlcm/hr.
2 at \4 / 2
Problem Solving for Chapter 2
Let {a, a^) and {b, - fc- + 2i> - 5) be the points of tangency.
For y = x^,y' = 2x and for y = —x^ + 2x-5,y'= - 2r + 2.
Thus, 2a= -2i) + 2=*a + i'= l,ora= 1 - b. Furthermore, the slope of the common
tangent line is
a"- - {-b^ + 2b - 5) ^i\ - by + b^ -2b + 5
a- b {\ - b)- b
\ -2b + Ir + b- -2b + 5
-2b + 2
2b
= -2b + 2
=>2b- - 4b + 6 = 4b^ - 6b + 2
=i' 2b- -2b- 4 = 0
=>/r-fo-2 = 0
=^(b- 2)(b + 1) = 0
* = 2, - 1
For b = 2, a=l— b=— I and the points of tangency are (— I, 1), (2, -5). The tangent line has slof)e —2:
y - 1 = -2(x = 1) ^y = -2r - 1
For b=—l,a=\—b = 2 and the points of tangency are (2, 4) and {- 1, — 8). The tangent line has slope 4:
y - 4 = 4(x - 2) ^ y = 4jt - 4.
4. (a) y = ^-, y ' = 2jc. Slope = 4 at (2, 4).
Tangent line: y - 4 = 4(x — 2)
y = 4x — 4
(b) Slope of normal line: — |.
Normal line: y - 4 = -jU — 2)
1 1 9
y = -4X + 2
y=-\x + l = x^^4x-+x-\S = 0=^ (4.x + 9)^ - 2) = 0
j: = 2, — 4. Second intersection point: (— 3, ig)
(c) Tangent line: y = 0
Normal line: .r = 0
—CONTINUED—
374 Chapter 2 Differentiation
4. —CONTINUED—
(d) Let {a, a'), a ^ 0, be a point on the parabola y = x^. Tangent line at {a, a^) isy = 2a{x — a) + a^. Normal line at [a, a^) is
y = - -— (x — a) + cP-. To find points of intersection, solve
2a
x^ = -^Jc - a) -^ a^
x'- + ^x = a^ + l
2a 2
11,11
4a ~\ 4a
X + -— = a + —^=>x = a (Point of tangency)
4a 4a
x + -—=—\a + --\ =>x = —a
4a \ 4a) 2a 2a
!£■+ 1
2a
The normal line intersects a second time at x =
6. f(x) = a + b cos ex
f'{x) = —be sin ex
At (0, 1): a + ^ = 1 Equation 1
. /77 3\ , /c-n-\ 3 ^ . ^
At I —, - I: a + b cosi ~r] ~ '^ Equation 2
— iicsLnl — 1=1 Equations
From Equation \, a = \ — b. Equation 2 becomes {\ — b) + b cos \~7']~'^^=^~i' + b cos "T = T
From Equation 3, fc = ;^ — r. Thus -, — r- H ; — r cos — r = —
. eiA . etrX . eiA \4 ) 2
l-cos(f) = icsin(^
Graphing the equation g{c) = - c sin ( — - J + cos ( -— ) — 1, you see that many values of c will work.
13 3 1
One answer: e = 2,b = —-,a = — =^f(x) = - - - cos 2x
2 2 2 2
Problem Solving for Chapter 2 2nS
8. (a) by- = x3(a - x);a,b > Q
2 _ -^^(0 ~ x)
b^
Graph >>! =
V^(a — x)
and;y2
V;c3(a - x)
b
(c) Differentiating implicitly.
lb'^yy'= -ix-ia - x) - x^ = lax- - Ax^
, (3ax^ - Ay?) __ „
' ~ 2b^
— \j
iax^ = 4x3
3a = 4x
3a
fc^2
-mi'
-^)
21a^\
64 U
«)
/
270"
256b- ~^
y = ±
373a2
16ft
Two
. (3a
points: 1 — ,
3 73 a
16fc
W--
373 a
16ft
-)
10. (a)
3- =
1 =
dx
dt '
x>/3 => ^ =
dt
1(8)-V3|
= 12 cm/sec
. 1,-=.
,dx
dt
D =
dD
-{,-.
y^){2x
dx
dt
+ 2v
dy\
dt)
(b)
Vx^ + .V' =
dx^
dt
dt
7t= +
1
.V"
8(12) + 2(1)
^_
98
49
(b) a determines the x-intercept on the right: (a, 0).
ft affects the height.
764 + 4 768 717
cm/sec.
dy dx
^ y ., ^ de ^dt~^dt
(c) tan 0 = ^ => sec- 8 ■ —- = :;
x dt X-
From the trrangle, sec 6
68 „ dd 8(1) -2(12) -16 -4 ,.
^- "^"'^^ ^ = ,/68\ = -68" = "F '"'^'''
64
376 Chapter 2 Differentiation
12. £'W = lim ^^' + ^^ ~ ^^'
A.T->0 Ax
£(x)E(Ax) - Ejx)
= lim
A.t-»0 Ax
,. ^, JE{Ax) - 1
= lim E(x){-^
Ax^O V Ax
^, , ,. £(Ax) - 1
= £(x) hm -^
A.r->0 Ax
But, £'(0) = lim EM_im . ,i^ MM^ = 1.
A.v->0 Ax A.t->0 Ax
Thus, E'{x) = E(x)E'(Q) = E(x) exists for all x.
For example: E{x) = e*.
14. (a) v(t) = - Y' + 27 ft/sec
27
a(t) = — T- ft/sec^
27 27
(b) v(f) = — -t + 27 = 0 => — r = 27 ^> f = 5 seconds
S(5) = -^^(5)- + 27(5) + 6 = 73.5 feet
(c) The acceleration due to gravity on Earth is greater in
magnitude than that on the moon.
CHAPTER 3
Applications of Differentiation
Section 3.1 Extrema on an Interval 378
Section 3.2 Rolle's Theorem and the Mean Value Theorem . 381
Section 33 Increasing and Decreasing Functions and
the First Derivative Test 387
Section 3.4 Concavity and the Second Derivative Test .... 394
Section 3.5 Limits at Infinity 402
Section 3.6 A Summary of Curve Sketching 410
Section 3.7 Optimization Problems 419
Section 3.8 Newton's Method 429
Section 3.9 Differentials 434
Review Exercises 437
Problem Solving 445
CHAPTER 3
Applications of Differentiation
Section 3.1 Extrema on an Interval
Solutions to Even-Numbered Exercises
2. fix) = cos
fix) = ~2^"*Y
/'(O) = 0
/'(2) = 0
fix) = - 3xjx + 1
fix) = -3x
3
^ix + 1)-'/^
+ Jx + l(-3)
ix + l)-i/2[;t + 2ix + 1)]
-fix + l)-'/2(3x + 2)
/t-3 =0
6. Using the limit definition of the derivative,
\x\) - 4
lim ^«^ = lim (i-
X - 0 x->o-
j-»0
X
\x\)
= 1
lim ^«^ = lim (^
x->0+ X — 0 x-*0* X — 0
4
1
/'(O) does not exist, since the one-sided derivatives are
not equal.
8. Critical number: x = 0.
X = 0: neither
10. Critical numbers: ;c = 2, 5
X = 2: neither
X = 5: absolute maximum
12. gix) = x'^ix^ - 4) = ;r* - 4je
g'ix) = 4x3 - 8;c = 4^(^2 _ 2)
Critical numbers: x = Q,x = ± Jl
14. fix)
4x
x^+ 1
^ (;c^ + 1)(4) - (4;c)(2x) ^ 4(1 - ;c^)
^ ^'" ix'- + 1)2 ix^ + 1)2
Critical numbers: x = ±1
16. /(0) = 2 sec 0 + tan e, 0 < e < 277
/'(e) = 2 sec 0 tan 0 -I- sec^ 6
= sec 0(2 tan 0 -1- sec 0)
sec 6
2f^U
1
\C0S 0/ COS 0.
sec2 0(2 sin 0 -H 1)
Ttt IItt
On (0, 2Tr), critical numbers: 0 = —r, 0 = —r-
o 0
378
Section 3.1 Extrema on an Interval 379
18. fix) = ^^, [0, 5]
f'{x) = - => No critical numbers
Left endpoint: (0,-1 Minimum
Right endpoint: (5, 5) Maximum
20. f{x) = x^ + 2;c-4, [-1, 1]
fix) = 2x + 2 = 2U + 1)
Left endpoint: ( - 1 , - 5) Minimum
Right endpoint: (1, - 1) Maximum
22. fix) =x^ - \2x, [0, 4]
fix) = 3^2 - 12 = -iix^ - 4)
Left endpoint: (0,0)
Critical number: (2, - 16) Minimum
Right endpoint: (4, 16) Maximum
Note: X - - 2 is not in the interval.
26. y = 3 - |t- 3|,[-1,5]
From the graph, you see that f = 3 is a critical number.
Left endpoint: (-1,-1) Minimum
Right endpoint: (5, 1)
Critical number: (3, 3) Maximum
24.
gix) = VTx, [-
1,1]
Left endpoint:
— 1, — 1) Minimum
Critical number
(0,0)
Right endpoint:
(1,1) Maximum
28
hit) = ^, [3, 5]
h'(t)=7:^.
Left endpoint: (3, 3) Maximum
Right endpoint: (5,-1 Minimum
30. gix) = secx
g'ix) = secttanjc
Left endpoint:
6' I
E 2.
6' V3
r, 1.1547
Right endpoint: ( t> 2 I Maximum
Critical number: (0,1) Minimum
32. y^x--l- cos;c, [-1.3]
>>' = 2a: — sinjT
Left endpoint: (- 1, - 1.5403)
Right endpoint: (3, 7.99) Maximum
Critical number: (0, - 3) Minimum
34. (a) Minimum: (4, 1)
Maximum: (1,4)
(b) Maximum: (1, 4)
(c) Minimum: (4, 1)
(d) No extrema
36. (a) Minima: (-2. 0) and (2. 0)
Maximum: (0, 2)
(b) Minimum: (—2.0)
(c) Maximum: (0, 2)
(d) Maximum: (l. v^)
380 Chapters Applications of Differentiation
38. fix
2- x\ 1 < jc < 3
2 - 3;c, 3 < x < 5
Left endpoint: (1,1) Maximum
Right endpoint: (5, - 13) Minimum
."•"
\(3. -■')
\(5.-13)
40. fix) = , [0, 2)
2 — X
Left endpoint: (0,1) Minimum
'
(0.1)
42. (a) 3
(b) fix) = 3;cV3 - X, [0, 3]
fix)
Maximum: I 2, -
Minimum:
(0, 0), (3, 0)
x(^)(3-x)-i/2(-l) + (3-;c)>/2(l)
3:
= |(3-;c)-'/2(|)[-x + 2(3-x)]
^ 2(6 - 3x) ^ 6(2 - x) ^ 2(2 - j:)
~ 373 - X ~ 3V3 -X ~ V3 -X
Critical number: x = 2
/(O) = 0 Minimum
/(3) = 0 Minimum
/(2) = !
Maximum: 2
44. /(x) =
/W =
fix) =
/"W =
1
[1-
x^+ 1'
-2x
2
(;c2 + 1)
-2(1 - 3x2)
(x2 + 1)3
24jc - 24x3
(x2 + !)-»
Setting/'" = 0, we have x = 0, ± 1.
if"(l)| ~ 7 is the maximum value.
46. fix)
x^+ 1
[-1,1]
24x - 24x3
/"'W = / 2 -L 114 (^^^ Exercise 44.)
/<%) =
/'5)(x) =
(x^ + D"
24(5x^ - lOx^ + 1)
(x2 + 1)5
-240x(3x^ - lOx^ + 3)
(x2 + 1)6
[/'"•'(O)! = 24 is the maximum value.
48. Let/(x) = 1/x. /is continuous on (0, 1) but does not
have a maximum. /is also continuous on (— 1, 0) but does
not have a minimum. This can occur if one of the
endpoints is an infinite discontinuity.
\
V
Section 3.2 Rolle's Theorem and the Mean Value Theorem 381
50.
52. (a) No
(b) Yes
54. (a) No
(b) Yes
v-sin2e 77 377
dt
IS constant.
(by the Chain Rule)
^ V- cos 20 tie
\6 dt
In the interval [77/4, 3 77/4], d = 77/4, 377/4 indicate
minimums for dx/dt and 0 = 77/2 indicates a maximum
for (ic/iir. This implies that the sprinkler waters longest
when 6 = 77/4 and 377/4. Thus, the lawn farthest from
the spinkler gets the most water.
58. C=2x + ^0^^,l.x<300
C(l) = 300,002
C(300) = 1600
X-
2x2 = 300,000
x2 = 150,000
X = lOOyn == 387 > 300 (outside of interval)
C is minimized when x = 300 units.
Yes, if 1 < jr < 400, then x = 387 would minimize C.
60. fix) = W
The derivative of/ is undefined at every integer and is
zero at any noninteger real number. All real numbers are
critical numbers.
62. True. This is stated in the Extreme Value Theorem.
64. False. Let/(x) = a;^. x = 0 is a critical number of/.
g(x)=f(x-k)
= (x- kV- ■
X = i- is a critical number of g.
Section 3.2 Rolle's Theorem and the Mean Value Theorem
2. Rolle's Theorem does not apply to/(x) = cot(x/2) over
[77, 377] since/is not continuous at x = 277.
4. fix) = x(x - 3)
x-intercepts: (0, 0), (3, 0)
f'(x) -2x-3 = 0atx =
6. /(x) = -3xVx+ 1
x-intercepts: (-1.0), (0, 0)
fix) = -3xi(x + l)-'/2 - 3(x + 1)'/^ = -3(x + l)->/2(| + (x + i:
fix) = -3(x + l)-'/2(|x + l) = Oatx = -|.
382 Chapters Applications of Differentiation
8. fix) = x^ - 5x + 4, [1, 4]
/(l)=/(4) = 0
/is continuous on [1, 4]. /is differentiable on (1, 4).
RoUe's Theorem applies.
fix) = 2x-5
Ix- 5 = Q ^> x = -
c value: —
10. fix) = {x- i){x + 1)2, [- 1, 3] •
/(-l)=/(3) = 0
/is continuous on [— 1, 3]. /is differentiable on (- 1, 3).
Rolle's Theorem applies.
fix) = (x- 3)(2)(;c +\) + {x+ 1)2
= (x+ l)[2x - 6 + x+ \\
= (x+ \){ix - 5)
c value: —
11. fix) = 3 - |;c- 3|,[0, 6]
/(0)=/(6) = 0
/is continuous on [0, 6]. /is not differentiable on (0, 6)
since /'(3) does not exist. Rolle's Theorem does not apply.
16. fix) = cos X, [0, I-it]
/(0)=/(27r) = 1
/is continuous on [0, 2Tr]./is differentiable on (0, 2Tr).
Rolle's Theorem applies.
fix) = — sinj:
c value: tt
r2 -
■,[-1,1]
14. fix) =
/(-i)=/(i) = o
/is not continuous on [— 1, 1] since /(O) does not exist.
Rolle's Theorem does not apply.
18. fix) = cos 2x.,
2
'K'f]
/-
12
^i-S*A?
Rolle's Theorem does not apply.
20.
fix) = secx.
JL 2L
"4'4
/'
-f\
V2
/is continuous on [- Tr/4, Tr/4]./is differentiable on
(— 7r/4, tt/4). Rolle's Theorem applies.
fix) = secjctanjc
sec jc tan X = 0
x = 0
c value: 0
22. fix) =x- x^l^, [0, 1]
/(0)=/(1) = 0
/is continuous on [0, l]./is differentiable on (0, 1).
(Note: /is not differentiable ■Ax = 0.) Rolle's Theorem
applies.
/'W = 1 -
1
7>l/^
0
1 =
33/?
1
27
1 _ 73
27 9
/3
c value: ^ = 0.1925
Section 3.2 Rolle's Theorem and the Mean Value Theorem 383
24. X(x) = --sin-
/(-l)=/(0) = 0
[-1,0]
/is continuous on [- 1, 0]./is differentiable on (- 1, 0).
Rolle's Theorem applies.
1 TT
/(x) = ---cos- = 0
TTX 3
-— arccos— [Value needed in (— 1, 0).]
« -0.5756 radian
c value: —0.5756
28.
J
t'o .;
26. CW= 10(- + ^^
(a) C(3) = C(6) = y
(b)
^'« = iof-^^o^
1
= 0
x^ + 6.x + 9 X-
2x:2 - 6x - 9 = 0
6d
a 08
6 ±673 3 ±373
In the interval (3, 6): c = ^ "^.^ = 4.098.
30./(;c)= |.x-3|, [0,6]
/is not differentiable at x = 3.
32. f{x) = x{x^ — j: — 2) is continuous on [— 1, l] and
differentiable on (— 1, 1).
/(I) -/(-I)
1
l-(-l)
f\x) = 3.t2 - 2.t - 2 = - 1
){x - 1) = 0
1
34. f{x) = (x + l)/.v is continuous on [1/2, 2] and
differentiable on (1/2, 2).
/(2) -/(1/2) (3/2) - 3
2 - (1/2)
3/2
= -1
/V) = -^
XT = 1
c = 1
-1
384 Chapters Applications of Differentiation
36. fix) = j^ is continuous on [0, 1] and differentiable on
(0, 1).
/(l)-/(0) 1 -0
1-0 1
fix) = 3;c2 = 1
V3
= 1
X = ±-
In the interval (0, 1): c
V3
40. fix) = x - 2 sin j: on [- TT, ir]
(a)
tangent^
^y^^'^canl
^
'tangent
(b) Secant line:
,i fi2ILlflzA^IJ:J-A=■^
TT — (— V) 2tT
y — TT = l(x — tt)
y = X
38. fix) = 2 sin X + sin 2x is continuous on [0, it] and dif-
ferentiable on (0, tt).
= 0
/M-/(0)_0-0
TT — 0 TT
fix) - 2 cos X + 2 cos 2j: = 0
2[cosj: + 2cos2j: - 1] = 0
2(2 cos a: - l)(cosx + 1) = 0
1
cos X = —
cos X = —I
IT Sir
In the interval (0, tt): c = — .
(c) fix) = 1 - 2 cos jc = 1
cos X = 0
^ = -2' Airi"'
+ 2
Tangent lines: v -(— -2] = i(j:- —
y = X — 2
y = X + 2
42. /(x) = -jc^ + 4x3 + 8^2 + 5^ (0, 5), (5, 80)
80-5
(b) Secant line: y - 5 = 15(x - 0)
0 = 15x - >> + 5
fix) = -4x3 + 12x2 + 16x
/(5)-/(l)_,,
5-1
-4c3 + 12c2 + 16c = 15
0 = 4c3 - 12c2 - 16c + 15
c = 0.67 or c = 3.79
(c) First tangent line: y - fie) = m(x - c)
y - 9.59 = 15(x - 0.67)
0 = 15x - y - 0.46
Second tangent line: y - fie) = mix — c)
y - 131.35 = 15(x - 3.79)
0 = 15x - >' + 74.5
Section 3.2 Rolle's Theorem and the Mean Value Theorem 385
44. 5(?) = 200( 5
(a)
\ 2 + r/
5(12) - 5(0) _ 200[5 - (9/14)] - 200[5 - (9/2)] _ 450
12-0
(b) S'(t) = 200
12
(2 + tf
450
7
1
1
(2 + tY 28
2 + ? = 277
t = 2V7 - 2 = 3.2915 months
S'{t) is equal to the average value in April.
46. f{a) = f(b) and /'(c) = 0 where c is in the interval (a, ii).
(a) g{x) = fix) + k (b) gix) = fix - k)
gia) = gib) = fia) + k gia + k) = gib + k) = fia)
g 'ix) = fix) =* g '(c) = 0 g Xx) = fix - k)
Interval: [a, b] g 'ic + k) = /'(c) = 0
Critical number of g: c . Interval: [a + A, i? + ^]
Critical number of g: c + k
■fia)
(c) gix) = fikx,
4f ) = ^(f
g'ix) = kf'ikx)
g'{j\ = kf'ic) = 0
Interval:
a b
k' k
Critical number of g:
48. Let Tit) be the temperature of the object. Then 7(0) = 1500° and r(5) = 390°. The average temperature over the
interval [0, 5] is
390 - 1500
5-0
-222°F/hr.
By the Mean Value Theorem, there exists a time to, 0 < t^ < 5, such that Ty^) = —222.
50. fix) = 3 cos^
(a)
-> '"■-^
fix) = 6 cos
TTXW TT
-37rcos^) sin
2 JJ\2
TTX
U>M
\M\ft
nf
jm
(b) /and/' are both continuous on the entire real line.
(c) Since /(- 1) =/(l) = 0, Rolle's Theorem applies on
[- 1, 1]. Since /(I) = 0 and/(2) = 3, Rolle's
Theorem does not apply on [1, 2].
(d) lim fix) = 0
-t— >3
lim fix) = 0
386 Chapters Applications of Differentiation
52. /is not continuous on [—5, 5].
'l/x, x¥^0
Example: f(x) =
0, jc = 0
54. False. /must also be continuous and differentiable on each
interval. Let
56. True
58. Suppose/(.r) is not constant on (a, b). Then there exists Xi and ^Cj in {a, b) such that/(j:|) i= f[x^. Then by the Mean Value
Theorem, there exists c in {a, b) such that
. ^ f{Xo) - fix,)
f'ic) = "^^^ — ^-^-^ ^0. .
This contradicts the fact that/'(jc) = 0 for all x in (a, b).
60. Suppose/U) has two fixed points c, and Cj. Then, by the Mean Value Theorem, there exists c such that
f (c) = = = 1.
This contradicts the fact that/'(j:) < 1 for all x.
62. Let/W = cos x.f'is continuous and differentiable for all real numbers. By the Mean Value Theorem, for any interval [a, b\
there exists c in (a, b) such that
-fie)
fib)-f{a)
b- a
cos b - cos a . '
; = — sm c
b - a
cos b - cos a = (-sin c)ib — a)
\cosb — cos a I = |-sinc||ii - a\
Icos b - cos a\ < \b - a\ since | — sin c\ < 1.
Section 3.3 Increasing and Decreasing Functions and the First Derivative Test 387
Section 3.3 Increasing and Decreasing Functions and the First Derivative Test
2.
y=-{x+l)^
Increasing on: (-
oo,
-1
Decreasing on: (-
-1,
oo)
6
x^
^ x+1
4. f(x) =^-lx-^
Increasing on: {- 1, 0), (1, oo)
Decreasing on: (-oo, - 1), (0, 1)
Critical numbers: x = Q, —2 Discontinuity: jc = — 1
Test intervals:
-oo < X < -2
-2 <x < -\
-1 < X < 0
0 < a: < oo
Sign of /'W:
y' > Q
y' < 0
y' < 0
y' > Q
Conclusion:
Increasing
Decreasing
Decreasing
Increasing
Increasing on (-oo, -2), (0, oo)
Decreasing on (-2, - 1), (- 1, 0)
8. h{x) = 21x- X?
h'{x) = 21 - l>x- = 3(3 - x)(3 + x)
h\x) = 0
Critical numbers: x = ±3
Test intervals:
— oo < j: < —3
-3 <.T < 3
3 < v < CO
Signof;z'(^):
/i' < 0
h' >Q
/i' < 0
Conclusion:
Decreasing
Increasing
Decreasing
Increasing on (—3, 3)
Decreasing on (— oo, —3), (3, oo)
4
10. y = ;c + -
X
, {x - 2){x + 2)
Critical numbers: jc = ±2 Discontinuity: 0
Test intervals:
-oo < X < -2
-2 < .t < 0
0 < .t < 2
2 < .r < oo
Sign of y ':
y' > 0
y' < 0
y' < 0
y' > 0
Conclusion:
Increasing
Decreasing
Decreasing
Increasing
Increasing: (— oo, —2), (2, oo)
Decreasing: (-2,0), (0,2)
388 Chapter 3 Applications of Differentiation
12, f(x) = x^ + &C+ 10
fix) = 2x + 8 = 0
Critical number: x = -4
Test intervals:
-oo < X < -4
-4 < X < oo
Sign of/'U):
/' < 0
/'>0
Conclusion:
Decreasing
Increasing
Increasing on; (— 4, oo)
Decreasing on: (— oo, — 4)
Relative minimum: (—4,-6)
14. fix) = -{x^ + Sx+ 12)
fix) = -2x - 8 = 0
Critical number: x = —4
Test intervals:
-oo < X < -4
-4 < ;c < oo
Sign of /'(x):
/'>0
/'<0
Conclusion:
Increasing
Decreasing
Increasing on: (— oo, — 4)
Decreasing on: (-4, oo)
Relative maximum: (-4,4)
16. f{x) = jc^ - 6*2 + 15
fix) = 3x^ - \2x = 3x(x - 4)
Critical numbers: x = Q,4
Test intervals:
-oo < X < 0
Q < X < 4
4 < a: < oo
Sign of /'(jc):
/'>0
/'<o
/'>0
Conclusion:
Increasing
Decreasing
Increasing
Increasing on (— oo, 0), (4, oo)
Decreasing on (0, 4)
Relative maximum: (0, 15)
Relative minimum: (4, — 17)
18. fix) = ix + mx - 1)
fix) = Ixix + 2)
Critical numbers: x = —2,0
Test intervals:
-oo < x < -2
-2 <;c < 0
0 < ;t < oo
Sign of /'(x):
f'>0
/'<0
/'>0
Conclusion:
Increasing
Decreasing
Increasing
Increasing on: (— oo, —2), (0, oo)
Decreasing on: (—2,0)
Relative maximum: (-2,0)
Relative minimum: (0, - 4)
Section 3.3 Increasing and Decreasing Functions and the First Derivative Test 389
20. f(x) = y - 32r + 4
fix) = Ax^-7>2 = 4(jr3 - 8)
Critical number: x = 2
Test intervals:
-oo < X < 2
2 < X < CO
Sign of/'(x):
/'<o
/'> 0
Conclusion:
Decreasing
Increasing
Increasing on: (2, oo)
Decreasing on: (-oo, 2)
Relative minimum: (2, - 44)
22. f(x) = x^'^ - 4
/'W
±r-'/3
3" 3jc'''3
Critical number: j; = 0
Test intervals:
— oo < JC < 0
0 < JT < oo
Signof/'U):
/'<o
/'>o
Conclusion:
Decreasing
Increasing
Increasing on: (0, oo)
Decreasing on: (- oo, 0)
Relative minimum: (0, - 4)
24. fix) = {x- l)'/3
1
fix)
3{x - 1)V3
Critical number: x = 1
Test intervals:
-oo < A < 1
1 < .v < oo
Sign of/'U):
/'> 0
/'>o
Conclusion:
Increasing
Increasing
Increasing on: (—00,00)
No relative extrema
26. /(x) = |x + 3| - 1
F + 3| [-1, X <
Critical number: j: = — 3
-3
-3
Test intervals:
-00 < X < -3
-3 < .r < 00
Sign of/'(.r):
/'<0
/'>0
Conclusion:
Decreasing
Increasing
Increasing on: (—3, 00)
Decreasing on: (—00, —3)
Relative minimum: (—3, — 1)
•^W%.1
fix) ^^ ^ ^^^'^
- (;c)(l)
^^^^ (x +
1)^
Discontinuity: x =
-1
1
{x + \y
Test intervals:
-00 < x < -\
- 1 < .t < 00
Sign of/'(.r):
/'>0
/'>0
Conclusion:
Increasing
Increasing
Increasing on: (-00, - 1), (— 1, 00)
No relative extrema
390 Chapters Applications of Differentiation
30
/w =
x + 3
x^
x
+
3
x^
fix) =
1
x"
6
-{x +
x'
3.
Critical number: x
=
-6
Discontinuity:
X =
0
Test intervals:
-oo < j: < -6
-6 < X < 0
0 < JC < oo
Sign off'ix):
/'<o
/'>0
/'<0
Conclusion:
Decreasing
Increasing
Decreasing
Increasing on: ( - 6, 0)
Decreasing on: (— oo, —6), (0, oo)
Relative minimum: ( —6, — —
32. /W
/'W =
3;c -4
X- 2
(x - 2)(2x - 3) - jx^ -3x- 4)(1) x^ - 4a: + 10
(^ - 2)2
Discontinuity: x = 2
Increasing on: (— oo, 2), (2, oo)
No relative extrema
(x - 2Y
Test intervals:
-oo < X < 2
2 < X < oo
Sign of/'W:
/'>o
/'>0
Conclusion:
Increasing
Increasing
34. fix) = sin X cos x = - sin 2x, 0 < x < 27r
fix)
cos 2x = 0
„ . . , , 77 377 5t7 777
Critical numbers: x = — , — -, — -, — —
4 4 4 4
Test intervals:
0<x<^
4
77 377
4 4
377 577
4 4
577 777
4 4
-— < X < 277
4
Sign of/'(x):
/'>0
/'<0
/'>0
/'<0
/'>o
Conclusion:
Increasing
Decreasing
Increasing
Decreasing
Increasing
Increasing on: (o, Jj, (^, ^j, (^, 277)
„ . /77 377\ /577 777
Decreasing on: [-^,^)\-J,-^
Relative maxima:
77 1 \ /577 1
4'2/'V4'2
Relative minima:
377 1\ /777 1
2/'\4' 2
Section 3.3 Increasing and Decreasing Functions and the First Derivative Test 391
36. f{x) = ;-, 0 < X < Itt
1 + cos-'j:
.,, , cos x{2 + sinlt)
(1 + cos-x)^
Critical numbers: x = — , — —
2 2
Test intervals:
0<x<^
TT 377
y<x<Y
377
— < j: < 277
Sign of /'W:
/'>o
/'<o
/'>0
Conclusion:
Increasing
Decreasing
Increasing
Increasing on: (0, — ),(—, 277
Decreasing on:
77 377
2' 2
Relative maximum: | -r-. 1
Relative minimum:
377
38. f(x) = 10(5 - Jx'- - 3x + 16), [0, 5]
5(2x - 3)
(a) /'W =
v^
3.V + 16
(b)
(c)
5(2;c - 3)
vv
3,t + 16
Critical number: -v
(d) Intervals:
(»■!)
(I-)
fix) > 0
fix) < 0
Increasing
Decreasing
/is increasing when/' is positive and decreasing
when/' is negative.
40./(a-)=^ + cos ^,[0,477]
(a)/'W=|-|sm^
2ji 3« 4ji
(c)
1
r
sm
—
0
2
2
r
sm
2
1
.V
77
2
0
Critical number: .v = 77
(d) Intervals:
(0, 77) (77. 477)
fix) > 0 f\x) > 0
Increasing Increasing
/is increasing when/' is positive.
392 Chapters Applications of Differentiation
42. /(f) = cos^t - sin^r = -2 sin^f = g{t), -2 < t < 2
fit) = -4 sin f cos t = -2 sin 2f
/symmetric with respect to y-axis
zeros off. ±—
Relative maximum: (0, 1)
Relative minimum: ( — — ,
.i.if.-.
44. f(x) is a line of slope ~ 2 =>/'W = 2.
46. /is a 4" degree polynomial =>/' is a cubic polynomial.
48. /has positive slope
In Exercises 50-54,/'(ar) > 0 on (-oo, -4),/'(x) < 0 on (-4, 6) and/'(x) > 0 on (6, oo).
50. g{x) = 3f(x)-3 .. 52. g(x) = -f{x) 54. g{x) ^ f{x - \0)
. g'(x) = 3/'W .. g'ix) = -fix) g'ix) =f'{x - 10)
g'(-5) = 3/'(-5) >0 g'(0) = -f'(0) > 0 ■ g'(8)=/'(-2) < 0
56. Critical number: x = 5
/'(4) = -2.5 =>/is decreasing atx = 4.
/'(6) = 3 =>/is increasing at j: = 6.
(5,/(5)) is a relative minimum.
58. s{t) = 4.9(sin e)!^
(a) v(r) = 9.8(sin e)r speed = |9.8 (sin e)t\
(b) If 6 = Tr/2, the speed is maximum,
v(f) = 9.8 1.
60. C =
(a)
3f
27 + f3'
? > 0
f
0
0.5
1
1.5
2
2.5
3
C{t)
0
0.055
0.107
0.148
0.171
0.176
0.167
The concentration seems greater near t = 2.5 hours.
(b) 025
The concentration is greatest when t = 2.38 hours.
, , „, _ (27 + f3)(3) - (3f)(3r^)
^^ (27 + f3)2
_ 3(27 - 2f^)
(27 + ;3)2
C = 0 when ; = 3/ 4/2 = 2.38 hours.
By the First Derivative Test, this is a maximum.
Section 3.3 Increasing and Decreasing Functions and the First Derivative Test 393
62. P = IMx
P' = 2.44
X'
20,000
X
5000, 0 < ;c < 35,000
0
10,000
X = 24,400
Increasing when 0 < jc < 24,400 hamburgers.
Decreasing when 24,400 < j: < 35,000 hamburgers.
Test intervals:
0 < JC < 24,400
24,400 < X < 35,000
Sign of/":
P' > 0
P' < Q
M. R= VO.OOIT'' -47+100
0.0047-3 - 4
(a) R' =
IJO.OOIT* - 47-+ 100
7=10°,/? = 8.3666n
= 0
(b)
The minimum resistance is approximately
/?« 8.37a at r= 10°.
66. f(x) = 2 sin(3.i:) + 4 cos(3a:)
The maximum value is approximately 4 All. You could use calculus by finding /'(x) and then observing that the maximum
value of/occurs at a point where /'(jf) = 0. For instance, /'(0. 154) » 0, and/(0.154) = 4.472.
68. (a) Use a cubic polynomial
fix) = a-jX^ + a^x' + a^x + a^.
(b) fix) = 3ayX^ + 2a2X + a^
(0, 0): 0 = ao (/(O) = 0)
0 = a, (/'(0) = 0)
(4, 1000): 1000 = 64^3 + \6a, (/(4) = 1000)
0 = 48a3 + Sa. (/'(4) = 0)
(c) The solution is Qq = a, = 0, ^2
375
125
f/\ -125,, 375,
fix) = —f-x^ + —X'-
(d)
\
-(4, laxi)
A
lO.O) \
394 Chapters Applications of Differentiation
70. (a) Use a fourth degree polynomial /W = a^x^ + a^x^ + a^x'^ + a^x + Aq.
(b) f\x) = 4a^ + 3a^x^ + 2a.^ + fl;
(1, 2): 2 = a^ + a^ + a2 + a^ + Oq
0 = 4a^ + 3a3 + laj + Qj
(— 1,4): 4 = (34 — 03 + Oj — fli + Oq
0 = — 4a4 + Sflj — 2a2 + Oi
(3, 4): 4 = 81^4 + 27^3 + Qqj + 3a, + flo
0 = 108^4 + 27a3 + 6^2 + a,
(/(I) = 2)
(/'(I) = 0)
(/(-I) = 4)
(/'(-i) = o)
(/(3) = 4)
(/'(3) = 0)
(c) The solution is a^ = T, a, = —5, Oj
2' "4
/W = -ix^ + y+ \x^
3 , 23
(d)
(-1.4)
(3.4)
(1.2) \
/
\
72. False
Let h{x) = /UJgU) where/(x) = g(x) = ;c. Then
h{x) = X- is decreasing on (—00, 0).
74. True
If /(x) is an «th-degree polynomial, then the degree of
/'(x)isn - 1.
76. False. . •
The function might not be continuous.
78. Suppose /'(jc) changes from positive to negative at c. Then there exists a and b'ml such that/'W > 0 for all j: in {a, c) and
f'{x) < 0 for all X in (c, b). By Theorem 3.5, /is increasing on (a, c) and decreasing on (c, fc). Therefore, /(c) is a maximum of
/on (a, i>) and thus, a relative maximum of/.
Section 3.4 Concavity and the Second Derivative Test
2. y= -x^ + 3x^ - 2, y" = -6;c + 6
Concave upward: (— oo, 1)
Concave downward: (l,oo)
4. /U) =
2x+ 1'-^ {2a: + 1)3
Concave upward: (-ido, — j)
Concave downward: (— 5, oo)
6.y
^ {-3x^ + 40.j;3 + 135x), y" = -^ x(x - 2)(x + 2)
270
Concave upward: (-00, -2), (0, 2)
Concave downward: (-2, 0), (2, 00)
8. h(x) = ^ - 5x + 2
h'(x) = 5x^-5
h'\x) = 20^3
Concave upward: (0, 00)
Concave downward: (-00, 0)
Section 3.4 Concavity and the Second Derivative Test
395
10. y = X + 2 CSC X, (- 77, ir)
y' = 1 — 2 CSC ;c cot a:
y" = — 2cscx(-csc^j:) - 2 cot a:(-csc jrcot^)
= 2(csc-'jc + cscj:cot^x)
Concave upward: (0, it)
Concave downward: (-tt, 0)
12. f{x) = 2x^ -Zx- - \lx + 5
fix) = (>x^-6x- n
f"(x) =\lx-(i
f"{x) = 12j: - 6 = Owhen.r =
Test interval
-oo < x < \
5 < .r < oo
Sign of /"W
fix) < 0
fix) > 0
Conclusion
Concave downward
Concave upward
Point of inflection: (5. ~t)
14. f(x} = 2x^ - 8x + 3
f'(x) = 8x^-8
fix) = 24x- = Owhenx = 0.
However, (0, 3) is not a point of inflection since /"(jc) > 0 for all x.
Concave upward on (-00, 00)
16. fix) = .r^U - 4)
fix) = x> + 3.r2(x - 4) ■ ■
= x'[x + 3(x - 4)] = 4xHx - 3)
fix) = 4x^ + Mx - 3) = 4x[x + 2ix - 3)] = 12a:{x - 2) = 0
fix) = 12xU - 2) = 0 when jr = 0, 2.
Test interval
-00 < X < 0
0 < X < 2
2 < .V < 00
Sign of /"W
fix) > 0
fix) < 0
fix) > 0
Conclusion
Concave upward
Concave downward
Concave upward
18.
Points of inflection: (0, 0), (2, - 16)
fix) = x^x + 1, Domain: [— 1, 00)
fix) = W^U + 1)-'/^ + V7TT = :^j=
I 2vx + 1
fix)
6VxTT - (3x + l)ix + 1)-'/^ 3x + 4
4(x +1) ~ 4(x + 1)3/2
/"(x) > 0 on the entire domain of/ (except forx = - 1, for which /"(x) is undefined).
There are no points of inflection.
Concave upward on (— 1, 00)
X + 1
20. fix) = — 7=- Domain: jc > 0
fix) =
X- 1
2x^/2
nx) = ^
Point of inflection: 3
Test intervals
0 < -v < 3
3 < v < oc
Sign of /"(.r)
/" > 0
f < 0
Conclusion
Concave upward
Concave dow nw ard
V3
4./3
396 Chapter 3 Applications of Differentiation
3x
22. f(x) = 2 CSC — , 0 < X < 2tt
,,/ s , 3;c 3;c
fix) = -3cscycoty
9/ 3x 3x 3x\
f"(.x) = -| csc^ — + CSC — cot'^ — I T^ 0 for any x in the domain of/.
Concave upward: ( 0. ~^ ). I ~^. 2'7t
Concave downward: I — , — 1
No points of inflection
24. f(x) = sin jc + cos j:, 0 < jc < 27r
fix) = cos-t - sinx
fix) = — sinjc — cosj:
f"{x) = Owhen;c
377 Ttt
4 ' 4 ■
Test interval:
3-n-
3-77 777
77r
-— < X < 277
4
Sign of/"U):
fix) < 0
/"W > 0
/"W < 0
Conclusion:
Concave downward
Concave upward
Concave downward
Points of inflection: ( "Z- ^ )' ( "4^' ^
26. /W = ;c + 2 cos X, [0, 27r]
f'{x) = 1 — 2 sin X
/"(x) = — 2 cos X
/"(x) = Owhenx = ^,^
•' ^ ' 2 2
Test intervals:
0<x<f
77 377
2<^<T
3l7 377
2 <^< 2
Sign of/"(x):
/"< 0
/">o
/"< 0
Conclusion:
Concave downward
Concave upward
Concave downward
Points of inflection:
77 77\ (377 377
2' 2
2 ' 2
28. /(x) = x2 + 3x - 8
/'(x) = 2x + 3
/"(x) = 2
Critical number: x = -5
/"(-!) > 0
Therefore, (-|, -7-) is a relative minimum.
30. /(x)= -(x-5)2
fix) = -2(x - 5)
/"W = -2
Critical number: x = 5
/"(5) < 0
Therefore, (5, 0) is a relative maximum.
Section 3.4 Concavity and the Second Derivative Test 397
32. f(x) =x^ -9x- + llx
fix) = 3;c2 - 18x + 27 = 3U " 3)-
f"{x) = 6(x - 3)
Critical number: x = 3
However, /"(3) = 0, so we must use the First Derivative
Test./'(x) > 0 for all x and, therefore, there are no
relative extrema
34. g{x) = -\(x + 2)Hx - 4)2
,, , -(x-4)(x- \)ix + 2)
g U) = ^
g"{x) = 3 + 3x - 1^2
Critical numbers: x = —2, 1, 4
g"(-2) = -9 < 0
(- 2, 0) is a relative maximum.
^"(1) = 9/2 > 0
(1, — 10.125) is a relative minimum.
^"(4) = -9 < 0
(4, 0) is a relative maximum.
36. f(x) = Jx- + 1
Vjc- + 1
Critical number: j: = 0
/"W = (^2 + 1)3/2
/"(O) = 1 > 0
Therefore, (0, 1) is a relative minimum.
38. f(x)
fix)
X
X - 1
-1
U - 1)^
There are no critical numbers and x = 1 is not in the
domain. There are no relative extrema.
40. f(x) = 2 sin.r + cos 2v, 0 < .v < 27r
f'(x) = 2 cos ;t - 2 sin It = 2 cos x - 4 sin jt cos x = 2 cos .t:(l - 2 sin x) = 0 when x = —,—, —-, -r-.
o i 0 2
f"(x) = -2 sinx - 4 cos 2x
r(f)>o
/<t) < »
r(f ) > 0
Relalive maxima: (f.5).(^.|)
Relative minima: ( T- 1 ). ( ^. ~ 3
398 Chapters Applications of Differentiation
42. f(x) = x^Ve^^-, [- V6, V6]
3x(4 - x^)
(a) /U) = ,
V6 - AT-
f'(x) = 0 when ;c = 0, a: = ±2.
6(x* - 9x2 + 12)
■^"^""^ (6 - x2)3/2
/"(x) = 0 when x = ±
h - 733
(b) /"(O) > 0 => (0, 0) is a relative minimum.
/"(±2) < 0 =» (±2, 4V2) are relative maxima.
Points of inflection: (±1.2758, 3.4035)
44. fix) = VS sin X, [0, 27r]
(a) /'(x) = x/2x cos X +
Critical numbers: x = 1.84, 4.82
/"(x) = -VSsinx +
cosx cosx smx
V2x v2x 2xV2x
2cosx (4x^ + l)sinx
72x
2x72^
_ 4xcosx — (4x2 + l)sinx
2x>/2x
(b) Relative maximum: (1.84, 1.85)
Relative minimum: (4.82,-3.09)
Points of inflection: (0.75, 0.83), (3.42, -0.72)
(c)
The graph of/ is increasing when/' > 0 and
decreasing when/' < O./is concave upward when
/" > 0 and concave downward when/" < 0.
(c)
/is increasing when/' > 0 and decreasing when
/' < O./is concave upward when/" > 0 and
concave downward when/" < 0.
46. (a)
/' < 0 means/
decreasing
/' decreasing means
concave downward
(b)
/' > 0 means/
increasing
/' decreasing means
concave downward
48. (a) The rate of change of sales is increasing.
S" > 0
(b) The rate of change of sales is decreasing.
S' > 0, S" < 0
(c) The rate of change of sales is constant.
S'= C,S"= 0
(d) Sales are steady.
5= C,S' = 0,5"=0
(e) Sales are declining, but at a lower rate.
S' < 0. 5"> 0
(f) Sales have bottomed out and have started to rise.
S' > 0
50.
Section 3.4 Concavity and the Second Derivative Test 399
52.
54.
56.
(b) Since the depth d is always increasing, there are no
relative extrema./'U) > 0
(c) The rate of change of d is decreasing until you reach
the widest point of the jug, then the rate increases
until you reach the narrowest part of the jug's neck,
then the rate decreases until you reach the top of the
jug-
60. (a) f(x) = Vx
fix) = |x-V3
fix) = -ix-V3
Inflection point: (0, 0)
(b) /"W does not exist at ^ = 0.
62. f(x) = ax^ + bx- + ex + d
Relative maximum: (2, 4)
Relative minimum: (4, 2)
Point of inflection: (3, 3)
f'(x) = 2ax' + 2bx + c,/"(.v) = 6ax + 2b
/(2) = ia + Ab + 2c + d = A
/(4) = 64a + 16Z? + 4c + rf = 2
56a + 12fc + 2c = -2 => 28a + 6fc + c = - 1
/'(2) = \2a + Ab + c = 0. /'(4) = 48a + 8t + c = 0, /"(3) = 18a + 2fo = 0
28a + 6fc + c = - 1 18a + 2fc = 0
12a + 4fc + c = 0 \6a + 2b= -\
\6a + 2b =-\ 2a =1
a = ^, b = — T, c= 12, d = —6
fix)
tX
ix~ + IZt
400 Chapters Applications of Differentiation
(-1000. 60)
(1000.90)
B
(0. 50)
64. (a) lineOA: 3' = -0.06x slope: -0.06
line CB: y = 0.04;t + 50 slope: 0.04
f(x) = ax^ + bx- + CX + d
fix) = 3ax2 + 2bx + c
(- 1000, 60): 60 = (- 1000)3a + (1000)2fc - 1000c + d
-0.06 = (1000)2 3a _ 2000fc + c
(1000, 90): 90 = (1000)^0 + (1000)^^ + 1000c + d
0.04 = (1000)2 3a + 2OOO6 + c
The solution to this system of 4 equations is a = - 1.25 x 10"^ b = 0.000025, c = 0.0275, and rf = 50.
(b) J = - 1.25 X 10-V + 0.000025^2 + o.0275.)t + 50 (c)
66. 5 =
(d) The steepest part of the road is 6% at the point A.
5.75573 8.52ir2 , 0.654r
+ 0.99987, 0 < r < 25
10^ ID'S 10^
(a) The maximum occurs when T == 4° and S « 0.999999.
(b)
(c) 5(20°) - 0.9982
5 10 15 20 25 30
68. C = 2x +
300,000
C = 2 - ^^^ = 0 when x = 100715 « 387
By the First Derivative Test, C is minimized when
X = 387 units.
70.S^^^,t>0
(b) S'it)
13,000f
(65 + f2)2
_ 13,000(65 - 3r2) _
^ ^'' ~ (65 + t^y "
t = 4.65
5 is concave upwards on (0, 4.65), concave
downwards on (4.65, 30).
(c) 5 '(f) > Oforf > 0.
As t increases, the speed increases,
but at a slower rate.
Section 3.4 Concavity and the Second Derivative Test 401
y\
72. /W = 2(sinx + cosx), /(O) = 2
/'W = 2(cosx- sinx), /'(O) = 2
f%x) = 2(-sinA: - cosx), /"(O) = -2
/'iW = 2 + 2(jc - 0) = 2(1 + x)
P/W = 2
P2W = 2 + 2U - 0) + 5(-2)U - 0)2 = 2 + 2x - x2
Pj'U) = 2 - 2x
P^'W = -2
The values of/, P,, Pj- and their first derivatives are equal at x = 0. The values of the second derivatives of/ and P2 are equal
at X = 0. The approximations worsen as you move away from x = 0.
74. /(x) =
Ax) =
X - 1'
-(x+l)
2v^(x - 1)='
/(2) = 72
/'(2) =
372
272
•^ ^""^ 4.r3/2(x - 1)3 • -^ ^-^ 872 16
„,, /-^( 372 V 372 572
P,(x) = 72 + — (x - 2) = — X + — -
Pi Xx) =
372
P,(x) = 72 + ( -^ |(x - 2) +
(-¥>-)- K¥)<—»= = ^ ^ ^u -» ^ ^u - 2,
p ,, . 3^^2372
P2 (x) = ;— + ,^ (x - 2)
P2't^)
4
2372
16
16
The values of/, P,, P; and their first derivatives are equal at x = 2. The values of the second derivatives of/ and P; are equal
at X = 2. The approximations worsen as you move away from x = 2.
76. /(x) = x(x - 6)- =x^ ' 12x- + 36x
/'(x) = 3x- - 24x + 36 = 3(x - 2)(x - 6) = 0
f"(x) = 6x - 24 = 6(x - 4) = 0
Relative extrema: {2, 32) and (6, 0)
Point of inflection (4, 16) is midway between the relative extrema of/.
402 Chapters Applications of Differentiation
78. p(x) = ax^ + bx^ + ex + d
p'{x) = 3ax^ + 2bx + c
p'U) = 6ax + 2b
6ax + 2b = 0
b
■'^'Ta
The sign ofpXx) changes at x = —b/3a. Therefore, {-b/3a, p{—b/3a)) is a point of inflection.
M_/_^UJ^Ucf-AU. '" '^
3a) "V 27aV \9aV \ 3aj 27a^ 3a
Whenp(x) =x^ - 3x^ + 2,a= \,b= -3, c = 0, andd = 2
-(-3)
--T^ + d
Xn =
= 1
■°" 3(1)
2(-3P (-3)(0)
+ 2= -2-0 + 2 = 0
^0 27(1P 3(1)
The point of inflection of p(jc) = x^ — 3jr^ + 2 is (xq, y^) = (1, 0).
80. False. /(x) = 1/x has a discontinuity at x = 0.
82. True
y = sin(to)
Slope: y' = b cos{bx)
-b < y' < b (Assumed > 0)
84. False. For example, let f{x) = {x - 2Y.
Section 3.5 Limits at Infinity
2. fix) =
2x
v^?T2
No vertical asymptotes
Horizontal asymptotes: y = ±2
Matches (c)
4. f(x) = 2 + •
x^+ 1
No vertical asymptotes
Horizontal asymptote: y = 2
Matches (a)
6. fix) =
2x^ - 3x + 5
x'^+ 1
No vertical asymptotes
Horizontal asymptote: y = 2
Matches (e)
8. fix) =
2x^
X + 1
X
10°
10'
102
103
10*
105
106
fix)
1
18.18
198.02
1998.02
19,998
199,998
1,999,998
lim fix) = CO (Limit does not exist.)
Section 3.5 Limits at Infinity 403
10. f[x) =
8x
JjF-1,
X
10'
102
103
10^
W
10«
10^
fix)
8.12
8.001
8
8
8
8
8
lim f{x) = 8
12. f{x) = 4 + -
x^ + 1
X
lO"
10'
102
103
10*
10^
10^
fix)
5
4.03
4.0003
4.0
4.0
4
4
lim f{x) = 4
14. (a) /iW = ■'-^-^ = = 5.r - 3 + -
X X X
lim h(x) = CO (Limit does not exist)
x—*oo
c 3 7
5 - - + —
X X-
(b) h(x)
fix) _ 5x- - 3.r + 7
lim h(x) = 5
(c) /jU)
fix) ^ 5x- - 3x + l ^ 5 _ ^ 7
X^ X^ X x^ .t^
16. (a) lim .? , ^ = 0
(b) lira
3.r3
- 1
3 -
Ix
ix -
- 1
3 -
2;c-
Zx - 1
lim ftW = 0
18. (a) lim , , , . = 0
(b) lim
x^o= 4^2 + 1
5x3/2
(c) lim
x^oo 4.v3/2 + 1 4
5.r3/2
=c4yx + 1
oo (Limit does not exist)
in r 3.r3 + 2 3 1
20. lim -—-, T-^; = - = -
x-^ao 9x3-2x2 + 7 9 3
22. lim 4 + - = 4 + 0 = 4
jr-»cxi V Xj
/l 4\
24. lim -X ^ = - oo (Limit does not exist)
x-.a= \2 X-J
y 1
26. lim , = lim , , .- (for x < 0, x = - v/?)
•t->-°= Vx + (l/x)
-1
404 Chapters Applications of Differentiation
28. lim — ^|^= lim 3 + (1 A) (f„, ^ ^ Owphavp-.
x^-oc Vl + (1/x)
,„ ,. X ~ COS j: ,. / , cos a:
30. hm = lim 1
Jl->co X x-tao V X
=1-0=1
Note:
32. lim cos - = cos 0 = 1
x->oo \X/
cos X
lim = 0 by the Squeeze Theorem since
_1 <- '^0^^ ^ i
a: ~ X ~ X
34. /u:
3x
lim /(jc) = 3
lim /U) = -3
_, ,. 1 ,. tanr
36. lim X tan - = lim = lim
X ->oo X I ->o* t t ->o*
s'mt 1
t cos /
(1)(1) = 1
(Letjf = 1/r.)
Therefore, y = 3 and y
horizontal asymptotes.
= - 3 are both
38. lim {2x - V4x^ + 1 )
lim
X — »oo
{2x - J\x^ + Ij
2x + jAx' + 1
2x+ V4a;2 + 1
= lim
1
JT ^oo 2x + V4;c2 + 1
= 0
40. lim (3;c + J^x^ - ;c) = lim
X— > — oo
(-^^^'•1^^^
/ T
-°o 3;c - V9x^ — X
= lim
= lim
79?
(for JC < 0 we have x = — V? )
- ^x~
1
x^-^l + V9 - (l/jc) 6
42.
j:
IQO
10'
W
103
lo-*
10^
10«
/W
1.0
5.1
50.1
500.1
5000.1
50,000.1
500,000.1
lim
- xjx^ — X x^
+ xVx
1 x'
Limit does not exist.
+ xV^^
lim
x^
X ' -♦oo X'
+ x^fx
Section 3.5 Limits at Infinity 405
44.
X
10"
10'
102
103
10*
10^
106
fix)
2.000
0.348
0.101
0.032
0.010
0.003
0.001
lim — = 0
^-"^ x~Jx
46. jc = 2 is a critical number.
f\x) < Oforx < 2.
fix) > 0 for ;c > 2.
lim f{x) = lim /W = 6
For example, let/U)
0.lU-2)2+ 1
+ 6.
48. (a) The function is even: lim f(x) = 5
j:— >-oo
(b) The function is odd: lim f{x) — —5
50. y
Intercepts: (3, 0), ( 0, -
Symmetry: none
Horizontal asymptote: y = 1 since
lim
X - 2
1
lim
J:-»oo X ~ 2
Discontinuity: x = 2 (Vertical asymptote)
52. y^
2x
9-x^
Intercept: (0, 0)
Symmetry: origin
Horizontal asymptote: v = 0
Vertical asymptote: ;c = ±3
54. y
x^ -9
Intercept: (0,0)
Symmetry: y-axis
Horizontal asymptote: v = 1 since
x"
lim
1 = lim
,x^-9 ' x-Xiox^-9'
Discontinuities: x = ±3 (Vertical asymptotes)
Relative maximum: (0, 0)
J
L
406 Chapters Applications of Differentiation
56. v =
2x2
x^ + 4
Intercept: (0,0)
Symmetry: y-axis
Horizontal asymptote: y = 2
Relative minimum: (0, 0)
58. x^ = 4
Intercepts: none
Symmetry: y-axis
Horizontal asymptote: y = 0 since
4 4
lim —^ = 0 = lim
:->-<x>X x->ooX'
.2-
Discontinuity: x = 0 (Vertical asymptote)
60, y =
2x
1
Intercept: (0, 0)
Symmetry: origin
Horizontal asymptote: y = 0 since
lim
2x
0 = lim
2x
1 -X2
Discontinuities: x = ±1 (Vertical asymptotes)
62, y = 1 +
1
Intercept: (-1,0)
Symmetry: none
Horizontal asymptote: y = 1 since
lim (l +-) = 1 = lim (l +-
JT— >-oo\ x/ j: ->oc \ X
Discontinuity: x = 0 (Vertical asymptote)
64. y = 4n - ^
Intercepts: (±1,0)
Symmetry: y-axis
Horizontal asymptote: y = 4
Vertical asymptote: x = 0
66. y =
Domain: (-oo, -2), (2, oo)
Intercepts: none
Symmetry: origin
Horizontal asymptotes: y = ± 1 since
lim , .
X ->tx. ^x- - 4
1, lim
Vertical asymptotes: x = ±2 (discontinuities)
-1.
L
I I I I I > »
J-L J-4_t
Section 3.5 Limits at Infinity 407
68. fix)
EH^^[^
.:l„
E^r
)
.,, . U^ - \){2x) - ;c^(2r) -2x ^ , ^
/ W = (^,2 _ 1)2 = (^2 _ 1)2 = 0 when.t = 0.
.„. ^ ^ (;c^ - l)-(-2) + lr(2)U^ - l)(2;c) ^ 2(3.t-- + 1)
^ ^ ' (x^ - D" (x^ - 1)3
Since /"(O) < 0, then (0, 0) is a relative maximum. Since /"(jt) ^ 0, nor is it undefined in the domain of/, there are no points
of inflection.
Vertical asymptotes: ,i: = ± 1
Horizontal asymptote: y = 1
70. f(x)
1
1
x- - x - 2 {x + l)(.r - 2)
.,, , -{2x- 1) ^ . 1
/W = (,2 _,_ 2)2 = 0 when. = -.
/"W =
(x^ -x- 2P(-2) + (2t - l)(2)(;c^ -x- 2)(2x - 1)
(a-2 - X-2Y
6(;c^ - ;c + 1)
^
^
{x--x- ly
Since /"(t) < 0, then (5, —5) is a relative maximum. Since /"(x) =?^ 0, nor is it undefined in the domain of/, there are no
points of inflection.
Vertical asymptotes: x = —\,x = 2
Horizontal asymptote: y = Q
2(x5 + 3jc' - \)
fix) = ^-. ,J = 0 when. « 0.5321, -0.6527, -2.8794.
ix- + X + \y
/"(o) < 0
Therefore, (0, 1) is a relative maximum.
f"i-2) > 0
Therefore,
-2,-r
is a relative minimum.
Points of inflection: (0.5321, 0.8440), (-0.6527, 0.4491) and (-2.8794, -0.2931)
Horizontal asymptote: y = 0
(-0.6527,0.4491)
2 (0J321. 0.8440)
(-2.8794.-0.2931)
408 Chapters Applications of Differentiation
74. g(x) =
g'U) =
g'U) =
2x
2
(3;c2 + 1)3/2
-18;c
(3x2 + 1)5/2
No relative extrema. Point of inflection: (0, 0).
T, ■ , 2
Honzontal asymptotes: v = +—i=
No vertical asymptotes
s
^^
76. fix) = liE^ Hole at (0, 4)
fix)
4x cos 2j: — 2 sin 2x
There are an infinite number of relative extrema. In the interval
(-277, 2Tr), you obtain the following.
Relative minima: (±2.25, -0.869), (±5.45, -0.365)
Relative maxima: (±3.87,0.513)
Horizontal asymptote: y = 0 -
No vertical asymptotes
6
(b) fix)
x^ -2^2 + 2
2jc2
J^_2x^ _2_
2^2 2x2 + 2xi
1
X + 1
gix)
(c)
The graph appears as the slant asymptote y = — j-^ + 1-
80. lim 100
V,/V2 ->00
^-Kfe] = >o*-o]-
100%
82. >- =
3.351f2 + 42.46 U - 543.730
f
(a) 5
(b) Yes. lim y = 3.351
Section 3.5 Limits at Infinity 409
S4.S = J^,t>0
(b) Yes. lim 5 = ^ = 100
86. lim 4^ = lim ^Sf "17^1"
Divide p{x) and q{x) by .t".
Case 1: Hn < m: lim^-rr- = lim —
xa= q{x) X-
Case 2: If m = n: lim '—r-r = lim
b„ +
a. +
b„ +
0 +
bo
fffl — 1 yffl
+ 0 + 0 0
+ ■ • • + 0 + 0 b„
= 0
+ 0 + 0 a„
'1 ''o b„ + ■ ■ ■ + 0 + 0 b„
Case 3: If n > m: lim '-rr = lim
fc„ +
fc„ + • • • + 0
A_ + ^
= ±oo.
xT-
88. False. Let y^ = v^TT, then Vi(0) = 1. Thus, v, ' = 1/(2 V.t + l) and v, '(0) = 1/2. Finally, ■
3'/'=-^(7TTF5and.v,'t0)=4.
ht\. p = or- + bx + 1, thenp(O) = l.Thus,p'= lax + bandp'(O) = 5 ^> fc = 5. Finally, p"= laandpT^O) =
Therefore,
/'W =
(-l/Sy + (l/2)x+ 1, .r < 0
.VxTT, ;c > 0
and/(0) = 1,
f(l/2) - (I/4).r, x <<d ^ \
1 / / \ and/'(0) = -, and
I l/(2v^^rT), .V > 0 2
- 1/4 , ;c < 0 _ , 1
and/'tO) = — .
L-i/(4(x+ \y/~), x> 0 4
/"W =
/"(a:) < 0 for all real x, but/(jr) increases without bound.
410 Chapters Applications of Differentiation
Section 3.6 A Summary of Curve Sketching
2. The slope of/ approaches oo as j:— >0 , and approaches
— coasx—^0*. Matches (C)
4. The slope is positive up to approximately x = 1.5.
Matches (B)
6. (a) Xq,X2,x^
(c) Xi
(e) X2,Xj
(b) X2,X3
(d) j:,
8. y
x^ + 1
1 - JC2 (1 - x){x + 1)
(x2 + 1)2 {x^ + 1)2
0 when.x = ±1.
>'''=-^^ = 0when;. = 0,±73.
Horizontal asymptote: y = 0
y
y'
y"
Conclusion
- oo < .r < - 73
-
-
Decreasing, concave down
x=-73
73
4
-
0
Point of inflection
-73 < x < -\
-
+
Decreasing, concave up
x= -\
1
2
0
+
Relative minimum
-1 < a: < 0
+
+
Increasing, concave up
;c = 0
0
+
0
Point of inflection
0 < j: < 1
+
-
Increasing, concave down
;c= 1
1
2
0
-
Relative maximum
1 < a; < 73
-
-
Decreasing, concave down
x=73
73
4
-
0
Point of inflection
73 < X < oo
-
+
Decreasing, concave up
Section 3.6 A Summary of Curve Sketching 411
10. y =
y =
x^
+ 1
x^
- 9
-
IQx
{x^
_9)2
60(
x2 + 3)
U^ - 9)3
0 when x = 0
< 0 when x = 0
Therefore, | 0, —^1 is a relative maximum.
Intercept: (0, --
Vertical asymptotes: j: = ±3
Horizontal asymptote: y = \
Symmetric about y-axis
12. /(,)=^±2^,^2
X X
-2
f'(x) = ^r < 0 when x i- Q.
■' x~
fix) = J ^ 0
Intercept: (-2,0)
Vertical asymptote: x = Q
Horizontal asymptote: y = 1
5
4
3--
H — I — I — !-»-«
14. fix) =x + ^
^. . , 64 (x - 4)(x^ + 4j: + 16) ^ .
/ (;c) = 1 " ~J = ~3 = 0 when x = 4.
192
fix) = ^ > 0 if X ?^ 0.
Therefore, (4. 6) is a relative minimum.
Intercept; (-2 4/4, o)
Vertical asymptote: x = 0
Slant asymptote: y = x
,, ,, , x3 4x
16. / x) = -T— - = X + -^ -
X — 4 X — 4
/'(x) = •''"^•r ~,!?^ = 0 when X = 0. ±2V3
(x- - 4)-
/ (•'^) = "7"^ 7u~ = 0 when x = 0
(x- - 4)3
Intercept: (0. 0)
Relative maximum: (-2v^, -3v^)
Relative minimum: (2^3,3^3)
Inflection point: (0, 0)
Vertical asymptotes: x = ±2
Slant asymptote: y = x
412 Chapters Applications of Differentiation
18. y = = 2x - 1 +
y' = 2-
x-2
3
x-2
2x2-8^ + 5
(x - 2)3
{x - 2Y (x - 2)2
+ 0
= 0 when x =
4+ V6
U-Jl
Relative maximum: I , - 1.8990
Relative minimum:
Intercept: (0, -5/2)
2
4+76
, 7.8990 j
Vertical asymptote: x = 2
Slant asymptote: y = 2jc — 1
fi+v?
.-1.899)
H 1-^
-8 -4
■kzu
-4 /•' ^ 6
20. gU) = xV9 - X Domain: ;t < 9
3(6 - x)
&\x)
2V9 -;c
0 whenj: = 6
^"W = 4(9^ _ ^)3/2 < 0 whenx = 6
Relative maximum: (6, 6V3)
Intercepts: (0, 0), (9, 0)
Concave downward on (- oo, 9)
1 (6. 6V3 )
22. y = xJXfs - x^ Domain: -4 < x < 4
2(8 - x2)
y =
716"
= 0 whenx = ±272
„ 2x{x-^ - 24) ^ . ^
>" = (16 _ ^2)3/2 = 0 when;: = 0
Relative maximum: (272, 8)
Relative minimum: (-272,-8)
Intercepts: (0, 0), (±4, 0)
Symmetric with respect to the origin
Point of inflection: (0, 0)
24. y = 3(x - 1)2/3 _ (^ _ 1)2
2 2 - 2(;c - l)''/3
y' ^ (x - l)'/3 " ^^"^ " ^^ = (^ 1 1)1/3 = 0 whenx = 0,2
(y 'undefined for X = 1)
^"=3(7^^^"^^°^°'^"''^^
Concave downward on (-oo, 1) and (1, oo)
Relative maximum: (0, 2), (2, 2)
Relative minimum: (1,0)
Intercepts: (0, 2), (1, 0), (- 1.280, 0), (3.280, 0)
Section 3.6 A Summary of Curve Sketching 413
26. y = -jU^ -3^ + 2)
y' = —x^-\- 1 =0 when x = ±\
y"= -2x = Owhenx = 0
y
y'
y"
Conclusion
-oo < x < -\
-
+
Decreasing, concave up
-t = -1
4
3
0
+
Relative minimum
-1 < * < 0
+
+
Increasing, concave up
x = Q
"3
+
0
Point of inflection
0 < X < \
+
-
Increasing, concave down
x= 1
0
0
-
Relative maximum
1 < j: < oo
-
-
Decreasing, concave down
28. fix) = iU - D' + 2
fix) = ix- 1)2 = Owhenjc = 1.
fix) = 2(;c - 1) = 0 when a; = 1.
fix)
fix)
fix)
Conclusion
-oo < X < \
+
-
Increasing, concave down
X = 1
2
0
0
Point of inflection
1 < j: < oo
+
+
Increasing, concave up
30. fix) = ix+ \)ix- 2)(;c - 5)
fix) = ix+ l)(.r - 2) + ix + 1)U - 5) + ix - 2)ix - 5)
= 3(x2 - 4;t + 1) = 0 whenjc = 2± VJ.
fix) = 6(.r - 2) = 0 when x = 2.
fix)
fix)
fix)
Conclusion
-oo < .X < 2 - yi
+
-
Increasing, concave down
jc = 2 - 73
673
0
-
Relative maximum
2 - V3 < x: < 2
-
-
Decreasing, concave down
X = 2
0
-
0
Point of inflection
2 < j: < 2 + VI
-
+
Decreasing, concave up
a; = 2 + 73
-673
0
+
Relative minimum
2 + 73 < .T < oo
+
+
Increasing, concave up
(2-^.6v^)
C+v^.-fr/J)
Intercepts: (0, 10), (- 1, 0), (2, 0), (5, 0)
414 Chapters Applications of Differentiation
32. y = 3.t^ - 6x^ + |
>>'= 12x3 - 12x = 12x(x2 - 1) = Owhenx = 0,jc = +1.
/3
y"= 36*2 - 12 = 12(3x2 - 1) = 0 whenx = ±^.
y
y'
y"
Conclusion
-oo < X < -1
-
+
Decreasing, concave up
x= -1
-4/3
0
+
Relative minimum
-l<x<-f
+
+
Increasing, concave up
V3
'- 3
0
+
0
Point of inflection
-f <x<0
+
-
Increasing, concave down
x = 0
5/3
0
-
Relative maximum
0<x<f
-
-
Decreasing, concave down
V3
^- 3
0
-
0
Point of inflection
f <X<1
-
+
Decreasing, concave up
x= 1
-4/3
0
+
Relative minimum
1 < X < oo
+
+
Increasing, concave up
34. /(x) = X* - 8x3 + ig^ _ i6x + 5
fix) = 4x3 - 24x2 + 36x - 16 = 4(x - 4)(x - 1)^ = 0 when x = 1, x = 4.
/"(x) = 12x2 _ 4g^ + 36 = I2(x - 3)(x - 1) = 0 whenx = 3, x = 1.
fix)
fix)
fix)
Conclusion
-oo < X < 1
-
+
Decreasing, concave up
x= 1
0
0
0
Point of inflection
1 < X < 3
-
-
Decreasing, concave down
x= 3
-16
-
0
Point of inflection
3 < X < 4
-
+
Decreasing, concave up
x = 4
-27
0
+
Relative minimum
4 < X < oo
+
+
Increasing, concave up
J 1(0. 5)
^ (I.O) 1(5.0)
(4. -27)
Section 3.6 A Summary of Curve Sketching 415
36. y=(x- ly
y'= 5{x- !)■' = Owhen;c= 1.
y"= 20U- 1)3 = Owhenx= 1.
>-
y'
y"
Conclusion
-oo < .I < 1
+
-
Increasing, concave down
X = 1
0
0
0
Point of inflection
1 < j: < oo
+
+
Increasing, concave up
I
(1.0)
1 2 3
38. y=\x^-6x + 5|
,^2{x- 3)(x^ - 6a: + 5) _ 2(x - 3)(x - 5){x - 1)
^ \x^ - 6x + 5| \(x - 5){x - 1)1
= 0 when jc = 3 and undefined when x = 1, .r = 5.
„ _ 2(x- - 6x + 5) _ Ijx - 5)(.r - 1
\x^ - 6x + 5| |(.r - 5)(;c - 1)
undefined when.x = 1, .r = 5.
12 3 4 5 6
v
y'
y"
Conclusion
-OO < .T < 1
-
+
Decreasing, concave up
x= 1
0
Undefined
Undefined
Relative minimum, point of inflection
1 < .V < 3
+
-
Increasing, concave down
.v = 3
4
0
-
Relative maximum
3 < X < 5
-
-
Decreasing, concave down
x = 5
0
Undefined
Undefined
Relative minimum, point of inflection
5 < j: < oo
+
+
Increasing, concave up
40. y = cos X - — cos 2x, 0 < x < Itt
y' = — sin ;c + sin 2j: = — sin .r(l — 2 cos x) = 0 when x = 0. tt,
y" = — cos j: + 2 cos 2x = -cos x + 2(2 cos-;r - 1)
77 Stt
y 3 "
= 4 cos^ X — cos j: — 2 = 0 when cos .r
1 ± V33
= 0.8431, -0.5931.
Therefore, x = 0.5678 or 5.7154, x = 2.2057 or 4.0775.
T, , • ■ /•^ 3\ /57r 3\
Relative maxima: I T' 7 )> 1 ~T'' T I
Relative minimum: tt.
Inflection points: (0.5678, 0.6323), (2:2057, -0.4449), (5.7154, 0.6323), (4.0775, -0.4449)
416 Chapter 3 Applications of Differentiation
42. y = 2(x - 2) + cot X, 0 < ;c < 77
y = 2 — csc^ X = Q when x = —, — -
4 4
y " = 2 csc^ X cot .T = 0 when x = —
„ , . . /377 377
Relative maximum: ( — , — 5
Relative minimum: I T' ":7 ~ 3
Point of inflection: ( — , 77 — 4 1
Vertical asymptotes: x = Q, tt
44. y = sec^l — 1 - 2 tan
(f)
1, -3 < * < 3
'-^-<f)'"(f)(f)-^-'(f)(f)^»--^
Relative minimum: (2, — 1)
-5-4-3-2-1
(2.-1)
46. g(;c) = ATCOtX, — 277 < JC < 277
sin x cos JC — j:
g'W
sin^j:
p'(0) does not exist. But lim x cot j: = lim = 1.
jr->o ^-»o tan X
Vertical asymptotes: x = ±277, ±77
-"«pK-T-4(-f-»MI''MT-°
Symmetric with respect to y-axis.
Decreasing on (0, 77) and (77, 277)
48. fix) = 5
1
X - 4 X +
h)
/h
X = - 2, 4 vertical asymptote
y = 0 horizontal asymptote
50. fix)
Ax
Jx^ + 15
y = ±4 horizontal asymptotes
(0, 0) point of inflection
52. /"is constant.
/' is linear.
/ is quadratic.
Section 3.6 A Summary of Curve Sketching 417
54.
(any vertical translate of/ will do)
56.
X
A
(any vertical translate of the 3 segments of/ will do)
58. If/'(jc) = 2 in [-5, 5], then/(.r) = 2x + 3 and/(2) = 7 is the least possible value of/(2). If/'(.r) = 4 in [-5, 5], then
f{x) = 4.t + 3 and/(2) = 1 1 is the greatest possible value of /(2).
60. g{.x)
3^ - 5x + 2,
x^ + 1
Vertical asymptote: none
Horizontal asymptote: v = 3
The graph crosses the horizontal asymptote v = 3. If a
function has a vertical asymptote at x = c, the graph
would not cross it since /(c) is undefined.
62. g{x) =
x^ + x-2
x - 1
{x + 2)(jc - 1) _ \x + 2.
if .r ^ 1
[Undefined, if.r = 1
X - 1
The rational function is not reduced to lowest terms.
hole at (1,3)
64. g(x) = ; = 2.V + 2 -
X - 5
X- 5
t
]{
T
The graph appears to approach the slant asymptote
y = 2x + l.
418 Chapters Applications of Differentiation
66. f(x) = tan(sin ttx)
(a) 3_
(c) Periodic with period 2
(e) On (0, 1), the graph of/ is concave downward.
(t>) fi—x) = tan{sm(- ttx)) = tan(-sin irx)
= — tan(sin ttx) = —f{x)
Symmetry with repect to the origin
(d) On (— 1, 1), there is a relative maximum at [\, tan l)
and a relative mmimum at (—5, -tan l).
68. Vertical asymptote: x = —3
Horizontal asymptote: none
x^
^ x + 3
70. Vertical asymptote: x = 0
Slant asymptote: y = —x
y = — X +
1 \ - x^
72. fix) = |M2 - (ax) = ^{ax)(ax - 2), a # 0
f'(x) = a^x — a = a(ax — 1) = 0 when x = —.
a
■ f"{x) = a^ > Oforall;c.
(a) Intercepts: (0,0), (-0
Relative minimimi: (~ ~t
Points of inflection: none
(b)
74. Tangent line at P: y - yQ= f'(xQ){x - x^
(a) Let>' = 0: -Jq = f'{xQ){x - x^)
f'(Xf)x = xJXxq) - .Vo
x = x -^^ = x -^^
° f'ixo) ""^ fix,)
x-mtercept: \Xg
fM
_ jy-^n.
fW
, 0
(c) Normal line: y - yo= ~j;7~\ix " ^0)
Let y = 0: -^q
■M
Z'Uo)
-^o/'W = -x + Xq
x = Xq + yof'ixo) = Xo+ fixo)f'{xo)
;t-intercept: (xq + /(xo)/'^. 0)
(e) \BC\ =
/fa)
fix.)
fix,)
(g) \AB\ = |a:o - fa +/fa)/'fa))| = |/fa)/'fa)|
(b) Letx = 0: >> - yo =/'fa)(~-*o)
3' = >'o - ^o/'fa)
>" = /fa) - -To/'fa)
J- intercept: (0,/fa) - xj'ix,))
(d) Let j: = 0: y - j-q = 7^("-*o)
>' = >'o +
/'fa)
y-intercept: ^0, y, + t^J
m |pc|2-,2 + /'/falU/fa)irfa)i+/(xa)!
iPCp =
[ffa)Vl + \fix,)f
/'fa)
(h) |APp=/fa)V'fa)- + V
\AP\ = |/fa)|Vl + [/'fa)]2
Section 3. 7 Optimization Problems 419
Section 3.7 Optimization Problems
2. Let X and y be two positive numbers such that x + y = S.
P = xy = x(S - x) = Sx- x^
-r- = S - Ix = 0 when x = -.
dx 2
-r^ = - 2 < 0 when Jc = -.
dx^ 2
P is a maximum when x — y = S/1.
4. Let X and y be two positive numbers such that xy = 192.
192
S = x + 2,y = ^- + 7,y
dS
0 when y = 8.
192
dy y-
d^S 384
S is minimum when y = 8 and x = 24.
-rrr = — 5- > 0 when y = 8.
6. Let X and y be two positive numbers such that
j: + 2v = 100.
P = xy = y(100 - 2y) = lOOy - If-
dP
dy
d^P
dy^
P is a maximum when x = 50 and y = 25.
100 - 4y = Owheny = 25.
= -4 < 0 when V = 25.
8. Let -t be the length and y the width of the rectangle.
2;c + 2y = P
P-2x P
A = xy = x\- - xj = —X - x-
dA P P
-— = — — 2r = 0 when x = — .
dx 2 4
d-A ^ ^ , P
—rr = — 2 < 0 when x = —.
dx- 4
A is maximum when x = y = P/4 units. (A square!)
10. Let X be the length and y the width of the rectangle.
xy = A
A
P = lx + 2y = 2x + 2[- j = 2x + —
dP ^ 2A ^ ^ r-
-— = 2 ;- = 0 when x = VA.
d\ AT
d^P AA
> 0 whenx = Va.
dx^ x^
P is minimum when x = y = ^/a centimeters.
(A square!)
420 Chapters Applications of Differentiation
12. fix) = Jx- 8, (2, 0)
From the graph, it is clear that (8, 0) is the closest point on
the graph of/ to (2, 0).
14. fix) = ix+ 1)2, (5, 3)
16. F
dF
dv ''
22 + 0.02v2
22 - 0.02v2
(22 + 0.02v2)2
= 0 when v = yiToO - 33.166.
By the First Derivative Test, the flow rate on the road is
maximized when v = 33 mph.
d= VU-
- 5)2 + [ix + 1)2 - 3]2
= ^jr-
- lOx + 25 + (;c2 + 2x - 2)2
= v^-
- lOx + 25 + y* + 4;c3 - 8x + 4
Vx* + 4^3 + ^2 - 18x + 29
Since d is smallest when the expression inside the radical
is smallest, you need to find the critical numbers of
gix) = ;c^ + 4a^ + x2 - 18;c + 29
g'ix) = Ax^ + 12x2 + 2x- 18
= 2(;c - l)(2x2 + 8;c + 9)
By the First Derivative Test, x = \ yields a minimum.
Hence, (1, 4) is closest to (5, 3).
18. 4jc + 3^ = 200 is the perimeter, (see figure)
: , , /200 - 4;c\ 8 ,^„ „
A = lxy = 2x( ^ 1 = -(50x - x^)
4^ = I (50 - 2a:) = 0 when.t = 25.
dx 3
d^A
dx^
16
< 0 when x = 25.
100
A is a maximum when jc = 25 feet and >" = "3" feet.
20. (a)
(c)
Height, X
Length & Width
Volume
1
24 - 2(1)
1[24 - 2(1 )]2 = 484
2
24 - 2(2)
2[24 - 2(2)]2 = 800
3
24 - 2(3)
3[24 - 2(3)]2 = 972
4
24 - 2(4)
4[24 - 2(4)]2 = 1024
5
24 - 2(5)
5[24 - 2(5)]2 = 980
6
24 - 2(6)
6[24 - 2(6)]2 = 864
dV
dx
(b) V = ;c(24 - 2;c)2, 0 < ;c < 12
(d) 1200
The maximum volume seems to be 1024.
d^V
dx^
d^V
dx^
= 2xi24 - 2jc)(-2) + (24 - 2jc)2 = (24 - 2x)(24 - 6x)
= 12(12 - x)i4 - x) = 0 when a; = 12, 4 (12 is not in the domain).
= 12(2jc - 16)
< 0 when x = 4.
When X = 4,V = 1024 is maximum.
Section 3.7 Optimization Problems 421
22, (a) P = 2x + 2irr
= 2x + 2^1)
= 2x + Try = 200
■f
» r
200 -2x 2
• >> = = —(100 - X)
(b)
Length, x
Width, y
Area, ry
10
-(100 - 10)
77
(10)- (100 - 10) == 573
TT
20
-(100 - 20)
TT
(20)- (100 - 20)= 1019
TT
30
-(100 - 30)
TT
(30)- (100- 30)- 1337
TT
40
-(100-40)
TT
(40)-(100 - 40) - 1528
TT
50
-(100 - 50)
IT
(50)-(100 - 50) - 1592
TT
60
-(100 - 60)
TT
(60)-(100 - 60) = 1528
TT
The maximum area of the rectangle is approximately 1 592 m^
(c) A= xy = x-(100 - ;c) = -(100;c - jc^)
TT TT
(e) 2000
Maximum area is approximately
1591.55 m2 (jc = 50 m).
24. You can see from the figure that A = xy and y =
A = x(^] = |(6;c - x').
6-x
dA 1
dx 2
dx"
(6 - 2j:) = 0 when x = 3.
= - 1 < 0 when x = 3.
1 ; 1 4 5 6
A is a maximum when x = 3 and y = 3/2.
422 Chapters Applications of Differentiation
26. (a) A = - base x height
= ^{2Vir=i5)(4 + h)
= Vl6 - h^{4 + h)
=^ = 4(16 - h'')-"\-2h){A + h) + (16 - h^y^
ah 2
= (16 - h'^Y^'\-h(A + h) + (16 - h^)]
-2(h^ + 2/! - 8) -2{h + 4)ih - 2)
Vl6 - h^
Vl6 - h^
dA
— - = 0 when h = 2, which is a maximum by the First Derivative Test.
ah
Hence, the sides are 2Vl6 — /i^ = 4V3, an equilateral triangle. Area = 12V3 sq. units.
4 + h j4~Th
(b) cos a
V8V4 + h
V8
tan a =
Vie
4 + /i
Area = 2(|j(v'l6 - ^^{4 + h)
- {4 + hYtana
= 64 cos" a tan a
A '{a) = 64[cos'' a sec^ a + 4 cos' (— sin a)tan a] = 0
^ cos" a sec- a = 4 cos' a sin a tan a
1=4 cos a sin a tan a
1 . ,
sin a = — => a = 30° and A = 12 V3.
(c) Equilateral triangle
28. A = 2ry = 2x^r'^ — x^ (see figure)
dA
dx
2{r^ - 2x^)
0 when x
72r
2 ■
By the First Derivative Test, A is maximum when the rectangle has dimensions
V2rby(72r)/2.
U 7:317
\
(-r. 0)
.^/P^]
(r.O)
Section 3. 7 Optimization Problems 423
30. xy = 36 => y = —
36
A = {x + 3)(v + 3) = U + 3)1 — + 3
= 36 + M + 3^ + 9
x
dA -108
Dimensions: 9x9
+ 3 = 0 => 3j:2 = 108 => X = 6, .V = 6
32. V = Trr-^/i = Vq cubic units or /i =
V„
5 = 2wr- + lirrh = 2( Ttr^ + -^
^ = lilirr - -^) = 0 when r= ^^ units.
dr \ r~ I \ 2ir
h =
Vn
Voi^nT-'' 2Vo
1/3
2r
77{ yV2^)2 77^/3 (2^)1/3
By the First Derivative Test, this will yield the minimum surface area.
34. V = Trr^x
x + 2TTr= 108 => X = 108 - 27rr (see figure)
V = 7rr-(108 - 27rr) = 77<108r- - 2m^)
dV
dr
7r(216r — Gvr-) = 6Trr(36 — irr)
= 0 when r = — and x = 36.
i2t/ "Si*
— -r = 77(216 - 127rr) < 0 when r = — .
dr- TT
Volume is maximum when x = 36 inches and r = 36/ tt = 1 1.459 inches.
36. V = TTX^h = 'TTx-\2^/r — jr) = 2itx-Jr~ — x~ (see figure)
f =4"'
i)(r- - x^y-(-2x) + 2a- Vr- - x^
2ttx
K2r2 - 3;c2)
= 0 when x - 0 and x- = -—- => x = — r— ■
3 3
By the First Derivative Test, the volume is a maximum when
X = -^ — and n = — ;=.
3 73
Thus, the maximum volume is
4Trr^
'--{Hih
73/ 373'
UT^^l
U-V^.
424 Chapters Applications of Differentiation
38. No. The volume will change because the shape of the container changes when squeezed.
40. V = 3000 = -Txr' + irr-h
, 3000 4
h = T - T'"
7rr' 3
Let k = cost per square foot of the surface area of the sides, then 2k = cost per square foot of the hemispherical ends.
C = 2k(4Trr^) + kilirrh) = k\ Strr^ + l-rrr
V
3000 4
Trr'
]-[
16 , , 6000
3 r
dr
^TTr - ^^1 = 0 when r = ^/-^ - 5.636 feet and /i = 22.545 feet.
.3 r^ ] V 2iT
By the Second Derivative Test, we have
dr^
32 12,000
^r-T + -, —
> 0 when r = ^i
'1125
277 •
Therefore, these dimensions will produce a minimum cost.
42. (a) Let x be the side of the triangle and y the side of the
square.
(b) Let X be the side of the square and y the side of the pen-
tagon.
A = -(cot^W2 + -|cot-jj>'- where 3x + 4y =20
^,.W5-U;o...^.
A' =
f'-(-J')H)=»
60
4V3 + 9
When x = 0,A = 25, when x = 60/(473 + 9),
A « 10.847, and when x = 20/3, A == 19.245. Area is
maximum when all 20 feet are used on the square.
(c) Let X be the side of the pentagon and y the side of the
hexagon.
A = jfcot ^|;i:2 + (cot ^\y^ where 5x + 6y = 20
= 5U^U,2(^)
20 - 5xV
'4' = |(cot|)j: + 3
^(-1
6
20 - 5x
0< x<A.
= 0
A = tIcoI^W^ + -lcot^Mwhere4;c + 5>' = 20
4 \2
= x^ + 1.72047741 4 - -^j , 0 < x < 5.
A'=2x- 2.75276384(4 - ^.x] = 0
X = 2.62
When.1 = 0,A = 27.528, when.x = 2.62, A = 13.102,
and when x = 5.A ~ 25. Area is maximum when all 20
feet are used on the pentagon.
(d) Let X be the side of the hexagon and r the radius of the
circle.
'?)^
A = -I cot — 1x2 + nr~ where 6x + 2'nr = 20
3v^ , ^ AO 3x\2 10
A'=3v^-6fi«-^) = 0
\ 77 77/
x = 1.748
;c = 2.0475
When ;c = 0, A = 28.868, when x - 2.0475,
/4 « 14.091, and when ;c = 4, -4 = 27.528. Area
is maximum when all 20 feet are used on the hexagon
When j; = 0,-4 = 31.831, when;c = 1.748, A = 15.138,
and when x = 10/3, A = 28.868. Area is maximum when
all 20 feet are used on the circle.
In general, using all of the wire for the figure with more
sides will enclose the most area.
Section 3.7 Optimization Problems 425
44. Let A be the amount of the power line.
A = h-y + Ijx- + /
dy
j^r+
= 0 when y
V3-
d^A 2;c2 ^^ X
T^ = T^; ;tt7; > 0 tor V = — ?=.
dy- {x- + ffl^ ■' 73
The amount of power line is minimum when y = j:/ 73 .
(-;t, 0) (.t. 0)
46. fix) = |;c2 ^W = i^;c^ - \x^ on [0, 4]
(a) 9
-^
(b) rfU) = f(x) - g(x) = 5X^ - [j^x* - \x^) = x^- YiX'^
d'{x) = 2x-\x^ = Q=^?,x = T'
=>;c = 0,272 (in [0,4])
The maximum distance is d = 4 when x = 2 V2.
(c) fix) = X, Tangent line at [ijl, 4) is
y - 4 = 272(;c - 2V2)
y = 2V2X - 4.
g'{x) = \ic' - X, Tangent line at (2V2, o) is
y-Q = (i(2v^)3 - 2ji){x - 272)
y = ijlx - 8.
The tangent lines are parallel and 4 vertical units apart.
(d) The tangent lines will be parallel. \i dix) = f(x) — g{x),
then d'{x) = 0 = f'(x) — g'(x) implies that/'(x) = g'(x)
at the point x where the distance is maximum.
48. Let F be the illumination at point P which is x imits from source 1.
0 when
F =
X-
U-,
id
-xT-
dF
dx
-2W,
x'
+
2U2
id - xy
/^..
X
r' id- xy
i/K~ d-x
id - x) Vfi = X l/T,
dVT,=x{i/T,+ VQ
di/T,
l/T,+ k/L
d^F 6W,
2
6W0
dx^ x^ id - xY
This is the minimum point.
> 0 when x
dl/T,
l/T,+ l/T.
5o.(a)r=-^^ + i^
dT
(b) :7r =
—CONTINUED—
dx 2jx~ + 4 4
.t \_
v^?Tl~ 2
Z\~ = .r= + 4
x^ = 4
x = 2
7T2) = V2 + ^ hours
- - = 0
426 Chapters Applications of Differentiation
50.
— CONTDWE
(c) r =
D—
^2
Jx^ + A
A
dT_
dx
X
1
= 0
V, ^X^ + 4 V2
X
V2
Vx2 + 4
sine =
V2
T
d depends on — only.
52.
Jx^ + d
^1
ll+V^r
V2
^)^
dT
X
+
X -
a
dx v,Vx2 + rfi2 VjVrf,^ + (a - x)-
Since
= 0
Jx^ + rfi^
we have
sin S] and , . ^ , , = - sin e.
Jdi + {a- xY
sin 01 sin St „ sin 6, sin 0-,
= 0 => = -.
V, V, V, V,
Since
d^T _ d{^
d^
dx^ V,(;c2 + rf,2)3/2 vld.2 + (^ _ ^)2]3/;
this condition yields a minimum time.
-> 0
(d) Cost = Jx^ + 4 C, + (3 - x)C2
7FT4 (3 - x)
(1/C,)
From above, sin 9 =
(I/C2)
1/Ci _ Q
i/Q c,
54. CW = Ikjx^ + 4 + A:(4 - x)
C'W =
Vx2 + 4
A: = 0
2x = V^^ + 4
4;c2 = jc2 + 4
3x2 = 4
2
73
Or, use Exercise 50(d): sin 0 = — = -
e = 30°
Thus, X
73-
56. V = ];TTr^h = ^nr^JXAA - r^
3 3
= |Jr2[|](144 - r2)-'/2(-2r) + 2rVl44 - rA
By the First Derivative Test, V is maximum when r = 4^6 and h = 4V3.
Area of circle: A = 7r(12)2 = 14477
Lateral surface area of cone: S = 'n{A^)J{A^Y + (473)^ = 48>/6i7
Area of sector: 14477 - 48^677 = - Or^ = 128
14477 - 48n/677 277/ f-:\ , , ^,
e = -—^ — = -—(3 - V6) " 1.153 radians or 66°
72 3 ^ '^
Section 3.7 Optimization Problems 427
58. Let d be the amount deposited in the bank, i be the interest rate paid by the bank, and P be the profit.
P = (0.12)rf - id
d = ki^ (since d is proportional to i^)
p = (o.i2)(w2) - i(ki^) = kio.m^ - (3)
^ = A:(0.24i - 2i^) = 0 when / = ^ = 0.08.
di 3
d^P
di
T^ = k(0.24 - di) < 0 when / = 0.08 (Note: k > 0).
The profit is a maximum when i = 8%.
60. P = -—s^ + 6s' + 400
dP 2 3
(a) — = -— J= + I2s = -— j(5 - 40) = 0 when a: = 0, i = 40.
ds 10 10
d-P ^ 3
di^ ~ 5
d^-P
s + n
_, , (0) > 0 =^ J = 0 yields a minimum.
d-P
-p;-(40) < 0 => i = 40 yields a maximum.
ds-
The maximum profit occurs when s = 40, which corresponds to $40,000 {P = $3,600,000).
(b)~= -Is +12 = 0 when s = 20.
ds^ 5 .
The point of diminishing returns occurs when 5 = 20, which corresonds to $20,000 being spent on advertising.
62. 5, = |4m - 1| + \5m - 6| + |10m - 3|
Using a graphing utility, you can see that the minimum occurs when m = 0.3.
Line ^ = 0.3^:
^2 = |4(0.3) - 1| + |5(0.3) - 6| + 1 10(0.3) - 3| = 4.7 mi.
428 Chapter 3 Applications of Differentiation
64. (a) Label the figure so that r^ = x^ + h^.
Then, the area A is 8 times the area of the region
given by OPQR:
h
y^^N
V
■wryiWv": : V
x :;
if
V-
%
V ii
:ii^ ^
^ ii
m y
= 8 he- + (jc - h)h
= ^\(r^ - x") + [x- >/;^^^)vr2rrpj
= &xVr^ - x^ + 4x2 - 4r2
A '(x) = 8vV— 12 -
8^2
8^2
7^
+ 8x = 0
Vr^
= 8x + 8Vr2 - x^
x'^ = xVr^ - ^2 + (r2 - x2)
2^2 - /^ = xVr^ - x~
4jc^ - 4xV + A^ = x^(r^ - x^)
5x^ - 5x^r^ + r^ = 0 Quadratic in x^.
Sf- ± 725/ - 20/^ _ r^t r^-i
10 - inP ± ^5J-
lO"-
Take positive value.
/TTT!
= 0.85065r Critical number
-2 v2
(c) Note that x^ = — (s + Vs) and r^ - x^ = — (s - Vs).
a u fi X
Co) Note that sin - = - and cos - = -. The area A of the
2 r 2 r
cross equals the sum of two large rectangles minus
the common square in the middle.
A = 2(2x){2^) - 4^12 = %xh - 4h^
= 8/^ sin - cos - — 4r^ sin^ -
2 2 2
A(x) = 8x7^2 - x2 + 4x2 _ 4^
= 8[^(5 + 75)^(5 - 75)]"' + 4^(5 + 75) - 4;-
= iro^^'^]
'''2 2 r-
+ 2r^ + -75r2 _ 4^
= 1^75 - 2.2 + ^^
= 2r2
5^--f
2.2(75 - 1)
= 4r2 sin 0 - sin2
A'(d) = 4r2(cos e - sin -cos ~ ) = 0
a a
COS 0 = sin - cos - = 2 sin 0
2 2
tane = 2
e = arctan(2) » 1.10715 or 63.4°
Using the angle approach, note that tan 0 = 2, sin 0 = — ^ and sin2| - 1 = — (1 - cos 6) = -\\
J.
75
Thus, A(e) = 4.21 sin 0 - sin2:
V75 A* V5
^ 4^^ - 1) , ,^(^ _ ^)
Section 3.8 Newton's Method 429
Section 3.8 Newton's Method
2. /U) = 2x2-3
fix) = 4x
x, = l
n
^n
/uj
/K)
X ^^'"^
" f'ix„)
1
1
-1
4
I
4
5
4
2
0.125
5.0
0.025
1.225
4. /(x) = tanjc
fix) = sec-j:
jci = 0.1
n
x„
/u„)
/'UJ
" /'UJ
I
0.1000
0.1003
1.0101
0.0993
0.0007
2
0.0007
0.0007
1.0000
0.0007
0.0000
6. /W = x5 + ;c - 1
fix) = 5.^+\
Approximation of the zero of/ is 0.755.
n
■l^n
/UJ
/'UJ
" fiXn)
1
0.5000
-0.4688
1.3125
-0.3571
0.8571
2
0.8571
0.3196
3.6983
0.0864
0.7707
3
0.7707
0.0426
2.7641
0.0154
0.7553
4
0.7553
0.0011
2.6272
0,0004
0.7549
8. fix) = x- 2jx + 1
/'W = 1 ^
Vx+ 1
Approximation of the zero of/ is 4.8284.
n
^.
/(^J
/'(xj
/'UJ
1
5
0.1010
0.5918
0.1707
4.8293
2
4.8293
0.0005
0.5858
.00085
4.8284
10. fix) =1-2x3
/'W = -&r2
Approximation of the zero of/ is 0.7937.
n
x„
/U„)
/'UJ
n-\)
fK)
" /'UJ
1
1
-1
-6
0.1667
0.8333
2
0.8333
-0.1573
-4.1663
0.0378
0.7955
3
0.7955
-0.0068
-3.7969
0.0018
0.7937
4
0.7937
0.0000
-3.7798
0.0000
0.7937
430 Chapters Applications of Differentiation
12. fix) = ^x'
3x
fix) = 2jc3 - 3
Approximation of the zero of/ is -0.8937.
Approximation of the zero of/ is 2.0720.
n
x„
/U„)
nx„)
fix,)
f'(x„)
X -^^^"^
" f'ixj
1
-1
0.5
-5
-0.1
-0.9
2
-0.9
0.0281
-4.458
-0.0063
-0.8937
3
-0.8937
0.0001
-4.4276
0.0000
-0.8937
n
J^.
fixj
fK)
fix,)
f'ixJ
X ^^'"^
" fix,)
1
2
-1
13
-0.0769
2.0769
2
2.0769
0.0725
14.9175
0.0049
2.0720
3
2.0720
-0.0003
14.7910
0.0000
2.0720
14. f{x) = x^ — cos X
fix) = 3x^ + slnjc
Approximation of the zero of/ is 0.866.
n
Xn
fix,)
fix,)
fix,)
fix,)
X ^^^"^
" fix,)
1
0.9000
0.1074
3.2133
0.0334
0.8666
2
0.8666
0.0034
3.0151
0.001 1
0.8655
3
0.8655
0.0001
3.0087
0.0000
0.8655
16. h{x)=f{x)-g{x) = 3-x-
x^+ 1
h'ix)
■1 +
Ix
(x^ + 1)2
Point of intersection of the graphs of /and g occurs
whenx« 2.893.
n
x.
hix,)
h'ix,)
hix,)
h'ix J
X ^^^"^
" h'ix,)
1
2.9000
-0.0063
-0.9345
0.0067
2.8933
2
2.8933
0.0000
-0.9341
0.0000
2.8933
18. hix) = fix) - gix) = x^
h'ix) = 2x + sin;ic
cosx
One point of intersection of the graphs of/ and g occurs
when X ~ 0.824. Since /U) = x^ and gix) = cos x are
both symmetric with respect to the y-axis, the other point
of intersection occurs when x ~ — 0.824.
n
X,
hix,)
h'ix,)
hix,)
h'ix,)
X '"^"""'^
""" h'ix,)
1
0.8000
-0.0567
2311 A
-0.0245
0.8245
2
0.8245
0.0009
2.3832
0.0004
0.8241
20. fix)=x"-a = 0
fix) = nx"-^
X,." - xn + a _ in - l)x," + a
Section 3.8 Newton's Method 431
_xl±J_
i
1
2
3
4
Xi
2.0000
2.2500
2.2361
2.2361
75 = 2.236
24. j:,,,=-^
2x:,3 + 15
3.r,'
I
1
2
3
4
.r,
2.5000
2.4667
2.4662
2.4662
Vl5 = 2.466
26. /(jc) = tan ;c
/'W = sec^jc
Approximation of the zero: 3. 142
n
x„
fiXn)
n\)
f'ixj
1
3.0000
-0.1425
1.0203
-0.1397
3.1397
2
3.1397
-0.0019
1.0000
-0.0019
3.1416
3
3.1416
0.0000
1.0000
0.0000
3.1416
28. y = 4x^-\2x^+l2x-3= f{x)
y'= 12x- - 24x + 12 = /'(x)
3
/'Uj) = 0; therefore, the method fails.
n
■^^
/UJ
/'UJ
/'UJ
" /'UJ
1
3
2
3
2
3
1
2
1
2
1
1
0
—
—
30. f{x) = 2 sin ;c + cos 2x
fix) = 2 cos X - 2 sin 2x
377
Fails because/'(.x,) = 0.
377
2
/UJ
/'UJ
32. Newton's Method could fail if f'{c) = 0, or if the initial value x^ is far from c.
34. Let g{x) = /U) - X = cot .r - jc
g'ix) = -csc^x - 1.
The fixed point is approximately 0.86.
n
x„
8{x„)
g'ix„)
H.vJ
^'U„)
, ^UJ
" ^'UJ
1
1.0000
-0.3579
-2.4123
0.1484
0.8516
2
0.8516
0.0240
-2.7668
-0.0087
0.8603
3
0.8603
0.0001
-2.7403
0.0000
0.8603
36. f(x) = sin X, f'(x) = cos x
(a)
2
(b) .t, = 1.8
x. = X, - ^r4 = 6.086
/Ui)
(c) A-, = 3
X, = X, - ^7^ - 3.143
/U,)
(d)
2-
; \ (1.8. 0.974)
1 ■
>OL (3,0.141)
. /'^Vy...^,_^l6.086. 0)
/
■'f'/V^
-1-
(3.143. 01 V-^
-2-
\
The .r-intercepts correspond to the values resulting
from the first iteration of Newton's Method.
(e) If the Initial guess .v, is not "close to" the desired zero
of the function, the .v-intercept of the tangent line may
approximate another zero of the function.
432 Chapter 3 Applications of Differentiation
38. (s0x„,,=x„{2-2x„)
(h) x„^i =xll- llxj
(■
1
2
3
4
^,
0.3000
0.3300
0.3333
0.3333
(
1
2
3
4
^i
0.1000
0.0900
0.0909
0.0909
\ = 0.333
40. f(x) = ;c sin jt, (0, it)
f'(x) = x cos X + sin jc = 0
Letting F{x) = /'(x), we can use Newton's Method as follows.
\F'{x) = 2 cos JC — a: sin x\
n = 0.091
n
^n
F{x^)
nxr)
X ^^'"^
" F'{x„)
1
2.0000
0.0770
-2.6509
-0.0290
2.0290
2
2.0290
-0.0007
-2.7044
0.0002
2.0288
Approximation to the critical number: 2.029
42. y=f{x)=x\ (4,-3)
d= J{x -AY + {y + 3)2 = V(x - 4)2 + (jc^ + 3)2 = Jx!^ + 7x^ - 8x + 25
rf is minimum when D = x^ + Ix^ — 8x + 25 is minimum.
g{x) = D'= 4x3 + 14a: - 8
g'ix) = 12x2 + 14
n
J^.
gixj
^'UJ
H^„)
X ^(^^^
1
0.5000
-0.5000
17.0000
-0.0294
0.5294
2
0.5294
0.0051
17.3632
0.0003
0.5291
3
0.5291
-0.0001
17.3594
0.0000
0.5291
X = 0.529
Point closest to (4, - 3) is approximately (0.529, 0.280).
(4, -3).
44. Maximize: C
C' =
3f2 + t
50 + t^
-2t* - 2f3 + 300r+ 50
= 0
(50 + r3)2
Let/(x) = 3/^ + 2«3 - 300; - 50
fix) = 12r3 + 6f2 - 300.
Since /(4) = -354 and/(5) = 575, the solution is in the mterval (4, 5).
Approximation: t = 4.486 hours
n
x„
fix J
f'ixj
f'ixj
X ^^'"^
" f'W)
1
4.5000
12.4375
915.0000
0.0136
4.4864
2
4.4864
0.0658
904.3822
0.0001
4.4863
Section 3.8 Newton's Method 433
46. 170 = O.SO&c^ - n.974x^ + 71.248;c + 110.843, 1 < jc < 5
Let/W = 0.808;c3 - 17.974jc2 + 71.248;c - 59.157
fix) = 2.424x2 _ 35,948;( + 71.248.
From the graph, choose x^ = I and Jt, = 3.5. Apply Newton's Method.
n
x„
/UJ
/'UJ
no
" /'UJ
1
1.0000
-5.0750
37.7240
-0.1345
1.1345
2
1.1345
-0.2805
33.5849
-0.0084
1.1429
3
1.1429
0.0006
33.3293
0.0000
1.1429
n
Xn
/UJ
f'U„)
fUn)
X -^^'"^
1
3.5000
4.6725
-24.8760
-0.1878
3.6878
2
3.6878
-0.3286
-28.3550
0.01 16
3.6762
3
3.6762
-0.0009
-28.1450
0.0000
3.6762
The zeros occur when x = 1. 1429 and x = 3.6762. These approximately correspond to engine speeds of 1 143 rev/min and
3676 rev/min.
48. True
50. True
52. f{x) = J A - x^ sin(jc - 2)
' Domain: [-2, 2]
x = -1 and x = 2 are both zeros.
fix) = J A - x- cosU - 2) -
Let ;c, = - 1.
V4 - X-
= sin(jr — 2)
n
Xn
/(.vj
/'UJ
" /'UJ
1
-1.0000
-0.2444
-1.7962
0.1361
-1.1361
2
-1.1361
-0.0090
-1.6498
0.0055
-1.1416
3
-1.1416
0.0000
- 1.6422
0,0000
-1.1416
Zeros: x = ±2,x= -1.142
434 Chapters Applications of Differentiation
Section 3.9 Differentials
2. fix) = - = 6x-^
fix) = - \lx-^
■12
Tangent line at 2
3 -12
-f (. - 2)
-" 2 2
X
1.9
1.99
2
2.01
2.1
f{x) =
6
1.6620
1.5151
1.5
1.4851
1.3605
T{x) =
3
-f
1.65
1.515
1
1.5
1.485
1.35
4. fix) = ^x
1
fix)
lj~x
Tangent line at (2, Jl):
y - fi2) = f'i2)ix - 2)
1
72
2^
U-2)
^ + 1
272 v^
x
1.9
1.99
2
2.01
2.1
f{x) =
-Ji
1.3784
1.4107
1.4142
1.4177
1.4491
T{x)-
->.^
1
72
1.3789
1.4107
1.4142
1.4177
1.4496
6. fix) = CSC ;c
fix) = -cscjTCOtj:
Tangent line at (2, esc 2): y - /(2) = /'(2)(;c - 2)
y — CSC 2 = (—esc 2 cot 2)(x - 2)
>; = (-CSC 2 cot 2)(.x; - 2) + CSC 2
x
1.9
1.99
2
2.01
2.1
fix) = cscx
1.0567
1.0948
1,0998
1.1049
1.1585
r(;c) = (-csc2cot2)(x-
- 2) + esc 2
1.0494
1.0947
1.0998
1.1048
1.1501
8. y = fix) = 1 - 2x\f'ix) = -Ax,x = Q,Lx = dx =
Ay = fix + Ax) - fix)
= /(-0.1)-/(0)
= [1 - 2(-0.1)2] - [1 - 2(0)2] = _o.02
-0.1
dy=f'ix)dx
= /'(0)(-0.1)
= (0)(-0.1) = 0
10. y=fix) = 2x+ \J'ix) = 2,x = 2,Ax = dx = 0.01
Ay = fix + Ax)- fix) dy = fix) dx
= /(2.01)-/(2) =/'(2)(0.01)
= [2(2.01) 4- 1] - [2(2) + 1] = 0.02 = 2(0.01) = 0.02
Section 3.9 Differentials 435
12. y = 3jc2/'
dy = Ix'^'^dx = -jj^dx
14. y = V9^^
tfy = -{9 - jc2)-'/2(-2x)<ic = Z^^t
2 V9 -;c2
16. y = v^ +
v^
-^ 'iv^ IxJ'x)''^ IxJ'x'^
18. y = j: sin a:
dy = U cos ;c + sin j:) (it
20. y
^^+ 1
rU^ + 1)2 sec^jftanx - sec^j:(2j:)1 ,
^^ = [ x^f-^. J^
[2 sec-j:(jr^tan j: + tan x — x)1 ,
U^TTp J^
22. (a) /(1. 9) =/(2 - 0.1) -/(2) +/'(2)(-0.1)
» 1 +(-l)(-0.1)= 1.1
(b) /(2.04) =/(2 + 0.04) =/(2) +/'(2)(0.04)
= 1 + (-1)(0.04) = 0.96
24. (a) /(1.9) =/(2 - 0.1) »/(2) +/'(2)(-0.1)
« 1 + O(-O.l) = 1
(b) /(2.04) = /(2 + 0.04) « /(2) + /'(2)(0.04) ,
= 1 + 0(0.04) = 1
28. (a) g(2.93) = ^(3 - 0.07) - g(3) + g'(3)(-0.07)
« 8 + 5(-0.07) = 7.65
(b) g(3.1) = g(3 + 0.1) = g(3) + g'(3)(0.1) ^
= 8 + 5(0.1) = 8.5
26. (a) 5(2.93) = ^(3 - 0.07) « g(3) + 5'(3)(-0.07)
- 8 + (3)(-0.07) = 7.79
(b) g(3.1) = g(3 + 0.1) = g(3) + g'(3)(0.1)
- 8 + (3)(0.1) = 8.3
30. A = \bh, i) = 36, /! = 50
db = dh = ±0.25
oW = ^fc d/z + U rfi
^A^dA = 3(36)(±0.25) + 3(50)(±0.25)
= ±10.75 square centimeters
32. j: = 1 2 inches
l^x = dx = ±0.03 inch
(a) V = x^
dV= 3x'dx = 3(12)2(±0.03)
= ± 12.96 cubic inches
(b) 5 = 6x2
dS= \2xdx= 12(12)(±0.03)
= ±4.32 square inches
34. (a) C = 56 centimeters
AC = rfC = ± 1 .2 centimeters
C
C = 2n-r :
A = Trr* = M::
77/ 47r
dA = -^CrfC = -^(56)(±1.2) = —
277 277 77
dA
33.6/77
A [1/(477)1(56)-
=^ 0.042857 = 4.2857%
dA (\/27T)CdC 2dC
^^T- (1/477)C= --^^0.03
^ , 0^ = 0.015 = 1.5%
436 Chapter 3 Applications of Differentiation
36. P = (500.x - x^) - i^-x^ - llx + 3000], x changes from 1 15 to 120
dP = (500 -2x- X + ll)dx = (577 - 'ix) dx = [577 - 3(115)](120 - 115) = 1160
Approximate percentage change: —(100) = (100) = 2.7%
38. V = f 777^, r = 100 cm, dr = 0.2 cm
AV = dV = Airr^dr = 4-77(100)2(0.2) = SOOOTrcm^
40. E = IR
R
^ _
E
I
dR = -~dl
R
-{E/ndi
dl
E/I
I
dR
R
=
dl
J
=
dl
I
42. See Exercise 41.
A = |(base)(height) = |(9.5cot e)(9.5) = 45.125 cot 6
dA = -45.125 csc^erffii
csc^ e de de
cot 6 sin 6 cos 6
^ 025^
(sin 26.75°)(cos 26.75°)
^ 0.0044
(sin 0.4669)(cos 0.4669)
= 0.0109 = 1.09% (in radians)
44. /z = 50 tan 9
9 = 71.5° = 1.2479 radians
dh = 50 sec^ d • dd
50 sec2(1.2479)
50tan(1.2479)
9.9316
■de
2.9886
■de
< 0.06
< 0.06
\de\ < 0.018
46. Let/(;c) = ^,x = 21,dx= - 1
fix + A;c) - fix) + /'(x)dc = ^ +
3^
etc
^^ =- ^7 + 3-i^(- 1) = 3 - ^ - 2.9630
Using a calculator, ,^26 = 2.9625
48. Let fix) =x^,x = 3,dx= -0.01.
fix + Ax) '-fix) ■+f'ix) dx = x' + 3x^dx
fix + Ax) = (2.99)3 = 33 + 3(3)2(-0.01) = 27 - 0.27 = 26.73
Using a calculator: (2.99)^ = 26.7309
Review Exercises for Chapter 3 437
50. Let/(jt) = tanjc,;c = 0, ^ = 0.05,/'(x) = sec^x
Then
/(0.05)»/(0)+/'(0)<it
tan 0.05 «= tan 0 + sec^ 0(0.05) = 0 + 1(0.05).
52. Propagated error = f{x + Aa:) - f(x).
relative error =
, and the percent error
X 100.
54.True,^ = ^ = a
Ax dx
56. False
Let/(j;) = -/x, X = \, and Ax = aLc = 3. Then
^y =f(x + Ax) -f(x) =/(4) -/(I) = 1
and
..=/W^ = ^(3) = f.
Thus, dy > Ay in this example.
Review Exercises for Chapter 3
2. (a) /(4) = -/(-4) =
(c)
-3
At least six critical numbers on (— 6, 6).
^■fi^)-^^,^^^^-\
fix) = X
1
+ (x2 + l)-'/2
(b)/(-3)= -/(3)= -(-4) = 4
(d) Yes. Since/ (-2) = -/(2) = -(-1) = 1 and
/(I) = — /(— 1) = —2, the Mean Value says that there
exists at least one value c in (—2. 1) such that
f\c)
/(l)-/(-2) -2-1
1 - (-2)
1 +2
= -1.
(e) No, lim/(x) exists because /is continuous at (0, 0).
j:-»0
(0 Yes, /is differentiable at x = 2.
6. No. /is not differentiable at .t = 2.
(x2 + 1)3/2
No critical numbers ■ -
Left endpoint: (0, 0) Minimum
Right endpoint: (2,2/^5) Maximum
8. No; the function is discontinuous at .r = 0 which is in the interval [—2, 1].
10. /(x) = - 1 < X < 4
fix) = --,
f(b)-f(a) _{\/A)- 1 _ -3/4
b - a 4-1 3
12.
f(x) = Vx - 2r. 0 < X < 4
"4
fix) =
2J-.-'
m
b -
-/(«)
- a
-6-0
4-0
/'(c) = -i
c
1 1
no = ^
c = 2
c = 1
438 Chapter 3 Applications of Differentiation
14. fix) = lx^-2,x+ \
fix) = Ax-3
f(b)-f(a) _2\ - 1 _
b - a 4-0
f\c) = 4c - 3 = 5
c = 2 = Midpoint of [0,4]
16. g(x) = (x + 1)3
g'U) = 3U + 1)2
Critical number: x = —\
Interval
-oo < x < -\
-\ < X < CO
Signofg'U)
g'(x) > 0
g'{x) > 0
Conclusion
Increasing
Increasing
18. f(x) = sinj: + cosjc, 0 < ;c < 2ir
/'U) = cos j: — sin j:
Critical numbers: x = —,x
4
5lT
4
Interval
0<x<^
4
TT Sir
4 4
— < X < 27r
Sign of fix)
fix) > 0
/'W < 0
fix) > 0
Conclusion
Increasing
Decreasing
Increasing
20. gix) = ^sm(f-l], [0,4]
^'U) = |(f cos(f -1
2 2
= 0 when x = 1 + -, 3 + —
TT IT
Relative maximum: (1 H — , -
2 3
Relative minimum: I 3 H — . — -
TT 2
Test Interval
0 < X < 1 +-
77
l+-<x<3+-
3 +- < X < 4
TT
Signofg'W
g'W > 0
g'ix) < 0
g'ix) > 0
Conclusion
Increasing
Decreasing
Increasing
22. (a) y = /I sin(Vfc7^f) + Bcos{Vk/mt)
y' = A ^/kjm cosy Jkjmt) - B^/kJmsiny^/kJmt)
s'm^k/mt A i r—r- \ A
••tan(VfeAwr) = -.
= 0 when
Therefore,
zosjkjmt B
sin( Vfc/m t) =
cosy Vk/mt) =
B
JA^ + B^'
When V = y ' = 0,
>1
y = A
vM^+S2
VA^ + 52
VA2 + 52
(b) Period:
27r
Frequency:
1
1
iTT/Vk/m 277
/k/m
Review Exercises for Chapter 3 439
24. fix) = (x + 2)Hx - 4) = x3 - lit - 16
fix) = 3;c2 - 12
fix) = 6x = Owhen^ = 0.
Point of inflection: (0, - 16)
Test Interval
-OO < AT < 0
0 < a: < OO
Sign of /"(or)
f"(x) < 0
fix) > 0
Conclusion
Concave downward
Concave upward
26. h(t) = t - 4jt+l Domain: [-l,oo)
9
h'{t)=l
h"(t)
Vr + 1
0 => f = 3
1
(/ + 1)3/2
1
/i"(3) = o > 0 (3, —5) is a relative minimum.
28.
12 3 4 5 6 7
30. C=(£)..(f),
x^ ^ 2
32. (a) S = -0.1222f3 + 1.3655f- - 0.9052r + 4.8429
fb) 35
(c) S'{t) = 0 when t = 3.7. This is a maximum by the
First Derivative Test.
(d) No, because the t^ coefficient term is negative.
Iv 2/v
34. hm - ., , ^ = lim . _^ ^ , , = 0
36. lim , .," = lim , , =
ar->oo Vx^ + 4 Jr-.oo Vl + 4/.t-
= 3
38. g{x)
x= + 2
lim
5.t=
= lim
.t ^oc X^ + 2 ;t -.OO 1 + (2/x2)
Horizontal asymptote: y = 5
= 5
40. f(x)
3x
lim . ., = hm , ., — =
^-»°o VAT + 2 ■'-►OO Vx- + 2/vj:
= lim
^-^0° Vl + (2/.v=)
= 3
lim
3j:
j:-»-oo ^x^ + 2 x-«-oc
3x/x
3
= lim — ^
X-----V1 +(2/.r^)
= -3
Horizontal asymptotes: v = ±3
440 Chapters Applications of Differentiation
42. fix) = \x^ - 3.r2 + 2x| = \x{x - l)ix - 2)|
Relative minima: (0, 0), (1, 0), (2, 0)
Relative maxima: (1.577, 0.38), (0.423, 0.38)
44. g{x) = — — 4 cos X + cos 2x
Relative minima: {Itrk, 0.29) where k is any integer.
Relative maxima: {(2k - 1)tt, 8.29) where k is any integer.
46. fix) = 4x^ - xf = x^i'i - x)
Domain: (—oo, oo); Range: (— oo, 27)
fix) = 12.x:2 - 4x^ = 4;c2(3 - x) = 0 when x = 0, 3.
f'U) = 2Ax - \2x- = 12t(2 - ;c) = 0 when.x: = 0, 2.
/"(3) < 0
Therefore, (3, 27) is a relative maximum.
Points of inflection: (0, 0), (2, 16)
Intercepts: (0, 0), (4, 0)
48. fix) = (x2 - 4)2 ■
Domain: (-oo, oo); Range: [0, oo)
fix) = Axix- - 4) = 0 when a- = 0, ±2.
fix) = 4(3.t- - 4) = 0 when;c = ±^.
/"(O) < 0
Therefore, (0, 16) is a relative maximum.
/"(±2) > 0
Therefore, (±2, 0) are relative minima.
Points of inflection: (±273/3, 64/9)
Intercepts: (-2, 0), (0, 16), (2, 0)
Symmetry with respect to y-axis
50. fix) = ix- 3)(;c + 2)3
Domain: (-cx), oo); Range: [-^^6^,oo)
fix) = ix- 3)(3)(;c + 2)2 + ix + If
= (4a; - l)ix + 2)2 = 0 when;c = -2,\.
fix) = iAx - l)i2)ix + 2) + (x + 2)2(4)
= 6(2;c - \)ix + 2) = 0when;c = -2,\.
Therefore,
16,875
■2se~) is a relative minimum.
Points of inflection: (-2, 0), (5,
Intercepts: (-2, 0), (0, -24), (3, 0)
m5\
16)
Review Exercises for Chapter 3 441
52. f(x) = (x- ly^^x + 1)2/3
Graph of Exercise 39 translated 2 units to the right (x replaces by j: - 2).
(— 1, 0) is a relative maximum,
(l, - l/i) is a relative minimum.
(2, 0) is a point of inflection.
Intercepts: (-1,0), (2,0)
54. fix) =
2x
1 +x'-
Domain: (—oo, oo); Range: [—1,1]
.,, , 2(1 - x)il + x) ^ . _^,
/ (x) = ,. . -^.^ = 0 when ;c = ± 1.
(1 + x-y
f"(x) = ~,^^^ ~.f •* = 0 when x = 0, + ^.
(1 + x-y
/"(I) < 0
Therefore. (1, 1) is a relative maximum.
/"(-I) > 0
Therefore, (— 1, — 1) is a relative minimum.
Points of inflection: (-^3, -^3/2), (0, 0), (73, 73/2)
Intercept: (0, 0)
Symmetric with respect to the origin
Horizontal asymptote: y = 0
56. f(x) =
Domain: (— oo, oo); Range:
"■{
(1 + X*){2x) - xMx>) ^x{l - x)(l + x){\ + X-)
/ W = — (TTTP = (TTl^p = 0 when.v = 0. ±1.
,,. ^ (1 + x^yq - 10:c^) -{2x- 2r^)(2)(l + .x^^Ax^) 2(1 - 12v^ + 3.t^) ^ .
/"(±1) < 0
Therefore, I ± 1, - 1 are relative maxima
/"(O) > 0
Therefore. (0, 0) is a relative minimum.
J 6 ± 733
Points of inflection
Intercept: (0,0)
Symmetric to the >'-axis
Horizontal asymptote: v = 0
.(.7^,o.4(.^
733
0.40
-I 1 f— NtH 1 (->
442 Chapters Applications of Differentiation
58. fix) = .r2 + - = ^^^^
X X
Domain: (-00, 0), (0, oo); Range: (-00,00)
/ U) = 2x - ^ = ^; = 0 when a; = ^.
/"W = 2 + ^ = ^^^^^^ = 0 when;c = - 1.
Airi^ > »
/ 1 3 \
Therefore, — ;=, —p^ is a relative minimum.
Point of inflection: (-1,0)
Intercept: (-1,0)
Vertical asymptote: x = 0
60. /(jc) = |;c- 1| + |;c- 3|
Domain: (-00,00)
Range: [2, 00)
Intercept: (0,4)
y
4 1[ (0, 4)
/
3-
2-
1-
v
J
/
—
h-
1
— H —
2
— 1 —
3
1 ^^
4
62. f(x) = —(2 sin ttj: - sin l-nx)
Domain: [- 1, 1]; Range:
-373 373
277 ' 277
f\x) = 2(cos TTx - cos Itm) = -2(2 cos im + l)(cos Trt - 1) = 0
Critical Numbers: x = +-, 0
f"(x) = 2it(— sin TTx + 2 sin 277x) = 277 sin 77x(— 1 + 4 cos •nx) = 0 whenx = 0, ±1, ±0.420.
2 -373^
By the First Derivative Test:
3' 277
2 3V3
is a relative minimum.
is a relative maximum.
v3' 277
Points of inflection: (-0.420, -0.462), (0.420, 0.462), (±1, 0), (0, 0)
Intercepts: (- 1, 0), (0, 0), (1, 0)
Symmetric with respect to the origin
64. f{x) = x", n is a positive integer.
(a)/'(x) = /j^-'
The function has a relative minimum at (0, 0) when n is even,
(b) f"{x) = nin- \)x"-^
The function has a point of inflection at (0, 0) when n is odd and n > 3.
Review Exercises for Chapter 3 443
f
66. Ellipse: frr + fz = 1' >" = tV144 - jc^
144 16 3
A = (It)! |Vl44 - X- 1 = ^;cVl44 - x'^
dA ^41" ^
dx 3LVl44 -a:-
_ 41" 144 - 2jc^
~ 3Lvi44-x-
= + yi44
^]
= Owhenx= 772 = 672.
The dimensions of the rectangle are Ix = 12^2 by y = -Vl44 — 72 = 4^2.
68. We have points (0, y), [x, 0), and (4, 5). Thus,
y- 5 5-0 5x
m = 7 = or V
0-4 A- x ■ X - 4
5x \2
Let/U) = l2 = ;c2 +
X - 4
fix) = 2;c + 50|
X- 4
X - 4 — X
. (x - 4Y
] = „
100^ n
X — -, -;t = 0
U - 4)3
■:i(x - 4)3 - 100] = Owhen.r = Oor^: = 4 + 3/lOO.
L =
V^
25.r=
(x - 4)2 X- 4
J{x - 4Y- + 25
yiOO + 4
yioo
7100-/3 + 25 = 12.7 feet
70. Label triangle with vertices (0, 0), (a, 0), and {b, c). The equations of the sides of the triangle are y = (c/b)x and
y = [c/(b — a)]ix — a). Let (x, 0) be a vertex of the inscribed rectangle. The coordinates of the upper left vertex are
(x, {c/b)x). The y-coordinate of the upper right vertex of the rectangle is {c/b)x. Solving for the .t-coordinate x of the
rectangle's upper right vertex, you get
c
7
— a
(b — a)x = bijc — a
b- a
-(x-a)
X =
a-b
X + a = a ; — X.
Finally, the lower right vertex is
Width of rectangle: a — x — x
(b.c)
^l.x-a)
(0.0) (.1,0) / (aO)
(o-ilf*..0)
Height of rectangle: -rx (see figure)
b
A = (Width)(Height) = [a- ^-j^x - ;cj(^.v] = (a - ^xj^x
dA I a \c I c
-Ix^V-'bTb^^^'
a\ ac lac „ , b
- r = T^x = 0 when .r = —
bl b kr 2
Al) = {" - \%%% = (f)(f) = r^ = %'''] = |(Area of triangle)
444 Chapters Applications of Differentiation
72. You can form a right triangle with vertices (0, y), (0, 0), and (x, 0). Choosing a point {a, b) on the hypotenuse
(assuming the triangle is in the first quadrant), the slope is
b b-0
■y
-bx
0 — a a — X
Let fix) = L^ = x^ + y^ = x^ +
fix) = 2x + 21
a — x
-bx \2
a - X
-bx
a — X
-ab 1
a - xy\
2x[{a - xy + ab^] . , . ^ .r-r^
— ^=^ -^-: = 0 when x = 0,a + </ab^.
(a - xy
Choosing the nonzero value, we have y = b + l/a^b.
L = J{a + i/^Y + {b + i/^y
= (a^ + 3a''''3&2/3 + 3^2/3^4/3 + ijiyr-
= (a2/3 + ^2/3)3/2 meters
74. Using Exercise 73 as a guide we have L, = a esc 9 and Lj = ^ sec d. Then dL/dd — — a esc 0 col 0 + b sec Oiaa 0 = Q when
. ^r-rr „ 7^2/3 + ^2/3 7^2/3 + ^2/3
tan 6 = </a/b, sec 6 = 1-775 , esc 6 = —^ and
^1/3
L — L^ + L2 = a CSC d + b sec 6 = a
This matches the result of Exercise 72.
,1/3
(a2/3 + j,2/3)l/2 (a2/3 + ^2/3)1/2
a'/3 + ^ /,l/3
(^2/3 + ^2/3)3/2^
76. Total cost = (Cost per hour)(Number of hours)
'2 ^/110\ llv 825
50 V
^^^•^«/Vv
dT^ 11^_ 825 ^ 11 v^ - 41,250
dv ~ 50 v2 ~ 50v2
= 0 when v = V3750 = 25 V^ « 61.2 mph.
d'^T 1650
dv^
0 when v = 25 V6 so this value yields a minimum.
78. fix) =x^ + lx+ \
From the graph, you can see that/(x) has one real zero.
fix) = 3^2 + 2
/changes sign in [— 1, 0].
n
Xn
/UJ
/'UJ
fi^n)
1
-0.5000
-0.1250
2.7500
-0.0455
-0.4545
2
-0.4545
-0.0029
2.6197
-0.0011
-0.4534
On the interval [-1,0]: ;c == -0.453.
Problem Solving for Chapter 3
445
80. Find the zeros of/(jc) = smirx + x — \.
f'(x) = TTCOS ITX + 1
From the graph you can see that/U) has three real zeros.
n
K
/U„)
/'UJ
/'UJ
X ^^'"^
1
0.2000
-0.2122
3.5416
-0.0599
0.2599
2
0.2599
-0.0113
3.1513
-0.0036
0.2635
3
0.2635
0.0000
3.1253
0.0000
0.2635
n
■\
/UJ
/'U„)
f'UJ
" /V„)
1
1.0000
0.0000
-2.1416
0.0000
1.0000
n
-"^n
fix J
/'UJ
nx„)
" fix J
1
1.8000
0.2122
3.5416
0.0599
1.7401
2
1.7401
0.0113
3.1513
0.0036
1.7365
3
1.7365
0.0000
3.1253
0.0000
1.7365
The three real zeros of/(.r) are a: = 0.264, x = 1, and x ^ 1.737.
82, y = V36 - x-
dx 2^ 736 -:<^
afy =
v'^
=5 (it
84. p = 75 - jx
4
Ap = ;7(8) - p(7)
^5-«Ul75
[Ap = o'p because p is linear]
Problem Solving for Chapter 3
2. (a) dV = 3x2a[x: = 3x2Ax
AV = (x + Ax)3 - ;c^ = 3.t2AA: + 3.r(A.t)= + (Ax)'
AV- dV = 3x{Axy + (AxY = [3xAx + (Ax)2]Ax
^ ^ ^
e
= eAjt, where e — > 0 as Ax — ) 0.
Av
(b) Let fi = -^ - /'(x). Then e^O as Ax->0.
Furthermore, Ay — dy = Ay - f'ix)dx = eAx.
446
Chapter 3 Applications of Differentiation
4. Let h{x) — g{x) — fix), which is continuous on [a, b] and
differentiable on [a, b). h(a) = 0 and h{b) = gib) - fib).
By the Mean Value Theorem, there exists c in ia, b)
such that
h'ic)
hib) - hja) gib) - fib)
b — a b — a '
Since h '(c) = g '(c) — /'(c) > 0 and b - a > 0,
gib) -fib) >0 ^ gib) >fib).
3a
One point of inflection.
6. (a) /' = lax + b,f" = 2a i= 0. No points of inflection.
(b) /' = 3ax^ - 2bx + c,f" =6ax + 2i = 0=> x =
(c) y' = kyiL — y) = kLy — IqP-
y" = kLy' - 2kyy' = ky'iL - 2y)
If y = —, then y" = 0 and this is a point of inflection because of the analysis below.
++++++
.V": 1-
d = JWn?, sin d = -.
a
Let A be the amount of illumination at one
of the comers, as indicated in the figure. Then
kl . „ klx
(13^ + x")
sin 0 ■
(132 + ;c2)3/2
(x2 + 169)3/2(1) - .r I I(x2 + 169)'/2(2x) "^
A'W = kl ^^^^, = 0
=^ (x^ + 169)3/2 = 3xHx^ + 169)'/2
x^ + 169 = 3x^
2x2 = 169
13
72
9.19 feet
By the First Derivative Test, this is a maximum.
Problem Solving for Chapter 3 447
10. Let T be the intersection of PQ and RS. Let MN be the perf>endicular to SQ and PR passing through T.
Let TM = X and TN = b - X.
SN MR
b - X X
SN = - — -MR
X
b - X X X
SQ = ^^^{MR + PM) = ^^^d
AW = Area = i^. + {(' " 'd)ib - .) = ^^d[x + ^' ' ""^'^
A'U)=^d
'x(4x - 2b) - [Ix- - 2bx + tr)'
[ x^ J
2x2 _ 2bx + b-
A \x) = 0 ^ 4^2 - Ixb = 2x- -2bx + Ir
2x^ = IP-
b
X = — 7=
v/2
Hence, we have SQ = d = 7—= — '-d = I v 2 -
X b/j2
\)d.
Using the Second Derivative Test, this is a minimum. There is no maximum.
S N Q
12. (a) Let M > 0 be given. Take N = ^/m. Then whenever x > N = Jm,
you have
f(x) = x~ > M.
(b) Let 8 > 0 be given. Let M
you have
Then whenever x > M
e .r'
1
(c) Let E > 0 be given. There exists N > 0 such that |/(.r) - L\ < e whenever .v > N.
Let 6 = — Let x = -.
N. y
If 0 < V < S = — , then - < — => .v > Af and
N X N
[fix) - L\
/(;)-^
448 Chapters Applications of Differentiation
14. Distance = J^^Tl? + 7(4 - xY + 4^ = f(x)
X 4 - X
f'(x)
0
74^ + x^ V(4 - x)2 + 42
xV(4 - x)2 + 42 = U - 4)742 + ;c2
x2[16 - 8a: + ^ + 16] = (x2 - fo + 16)(16 + ;c2)
32x2 _ 8^3 + j;" = x'' - 8;c3 + 32;c2 - 128x + 256
128x = 256
x = 2
The bug should head towards the midpoint of the opposite side.
Without Calculus; Imagine opening up the cube:
The shortest distance is the line PQ, passing through the midpoint.
16. (a) s =
v^^lOOO-^
hr \ km
3600
sec
hr
18
V
20
40
60
80
100
s
5.56
11.11
16.67
22.22
27.78
d
5.1
13.7
27.2
44.2
66.4
d{t} = 0.071^2 + 0.389i + 0.727
(c) 1°
T = -(O.O7I52 + 0.3895 + 0.727) + —
s s
The minimum is attained when 5 = 9.365 m/sec.
(b) The distance between the back of the first vehicle
and the front of the second vehicle is d{t), the safe
stopping distance. The first vehicle passes the given
point in 5.5/ s seconds, and the second vehicle takes
d(s)/s more seconds. Hence,
^ d{s) 5.5
s s
(d)
T(s) = 0.0715 + 0.389 +
6.227
T'{s) = 0.071
6.227
, ^ 6.227
■ ■^ 0.071
• 5 ~ 9.365 m/sec
r(9.365) « 1.719 seconds
9.365 m/sec • ^^ = 3.37 km/hr
(e) 49.365) = 10.597 m
18. (a)
X
0
0.5
1
2
1
1.2247
1.4142
1.7321
71 +x
f-
1
1.25
1.5
2
(h) Let f(x} = 71 + X. Using the Mean Value Theorem on the interval [0, x],
there exists c,0 < c < x, satisfying
1 /U)-/(0) _ 7m^- 1
271 + c X - 0 x
Ac) =
Thus 71 + X
— , + 1 < — 1-1 (because 7l + c > 1).
271 + c 2
CHAPTER 4
Integration
Section 4.1 Antiderivatives and Indefinite Integration 450
Section 4.2 Area 456
Section 4.3 Riemann Sums and Definite Integrals 462
Section 4.4 The Fundamental Theorem of Calculus 466
Section 4.5 Integration by Substitution 472
Section 4.6 Numerical Integration 479
Review Exercises 483
Problem Solving 488
CHAPTER 4
Integration
Section 4.1 Antiderivatives and Indefinite Integration
Solutions to Even-Numbered Exercises
dx\ X XT
dx\ 37^ J dx\3
x^ - 1
= tA/2 _ ^-3/2 = ± i
4. '^'
r= -nG^ C
Check: — [ttO + C] = 77
8.? = 2;c-3
ax
;>' = — — + C = — r + c
-2
Check:
+ C
= 2x-
Given
10.
Rewrite
J^^ : ]■
X ^cbc
Integrate
-1
+ C
Simplify
12. L(a:2 + 3) ^ I (jc3 + 3x) die j + sfyj + C jx'' + ^x^ + C
"•J(3^ i if
-^dx
I/JC"
9V-1
+ C
4 2
9x
+ C
/'
16. (5 - x)^ = 5x - y + C
Check: ^5^-Y + C'=5-a:
/
18. (4x3 + 6^2 _ 1)^ = ^4 + 2x3 - X + C
Check: -f [x" + 2x3 - x + C] = 4x3 +6x2-1
ax
/<
20. I (x3 - 4x + 2)rfx = — - 2x2 + 2x + C
Check: 4"
dx
x^
— - 2x2 + 2x + C
4
= x3 - 4x + 2
22./
^^ " ^) ^ = /(^"^ " r'') "^-Iti^ iQ ^ ^ - ¥'' - ^"' ^ ^
Check: -^f |x3/2 + x'/2 + c) = x'/^ + ix-'/2 = v^ + -^
dx\Z I 2 2Vx
450
Section 4. 1 Antiderivatives and Indefinite Integration 451
i. Uv^ + i)dx= I (.
24. I (i/x^ + l)dx= I (jc^/" + 1) dx = V/t +X+ C
Check: ^f^^^'''' + ;c + c) = .x^/" + 1 = V? + 1
26.
1^- = /
;c-''dx = — + C
Check: "rl "A + c) = ^
2x'
+ C
28. r'"^y — -dx = 1^-2 + 2x-3 - 3x-'')d:x 30. j (2/2 - 1)^ dt = j (4r' - 4?^ + 1) rfr
-1
X~' 2x~2 "Ir-J
-1 -2 -3
J + - + C
Check:
dx
X x'^ jc"
^ x-2 + 2x'^ - 3x-'*
x^ + 2x-2
X*
4. 4
f3 + ; + C
Check:
^(1,5 _ 1,3
dt\5 3
r^ + ; + c = 4r* - 4/2 + 1
(2/2 - 1)2
32. 1(1 + 3f)r2d; = j(;2
+ 3/3) rfr = |/3 + I'' + C
Check: ^(j/' + I/-* + C) = /2 + 3/' = (1 + 3/)/2
34. 3 A = 3/ + C
Check: —(3/ + C) = 3
dt
36. I (f2 - sin /) dt = -/^ + cos / + C
Check: — -/^ + cos / + C = /2 - sin /
dt\3 J
38.
(e- + sec2 e) (ie = -e^ + tan e + c
Check: — (-e^ + tan 6 + c] = 6- + sec2 e
40. sec ^(tan y — sec y) dy = (sec y tan y — sec2 y) dy
= sec y - tan y + C
Check: -;-(sec y — tan y + C) = sec y tan v — sec2y
dy
= sec y(tan y - sec y)
42. f-^^?iViv= f^iT= ff-^y^ix
J 1 — cos-x J sin-.x J \sin.x/Vsin.x/
= esc X cot X d.x = - CSC X + C
_, , li r ^T 1 cos X
Check: -H -esc x + CI = esc x cot x + — — • — —
dx sin X Sin x
cosx
1 — cos2x
44. f(x) = Jx
46. /'(x) = X
48. fix) =
fU)
+ C
452 Chapter 4 Integration
50. ^ = 2U - 1) = 2x - 2, (3, 2)
ax
= \2{x -
l)dx = x^ -2x+ C
2 = (3)2 - 2(3) + C => C = - 1
y = x^-2x-l
52.
dy
dx
y
= -x-2- (1,3)
/-
;c-2cic = - + C
3 = Y+C=> C = 2
v = - + 2, x>0
54. (a)
x"
1. (-1,3)
y=j-x+C
3=^-(-l) + C
3 = --+ 1 + C
C =
56. g'(^) = 6x2,g(0)= -1
g(0) = - 1 = 2(0)3 + C =* C ^
^(;c) = 2^3-1
58. f'is) = 6s - Ss\ /(2) = 3
fis) = \{6s - 8s^)ds = 3^2 - 2j'' + C
/(2) = 3 = 3(2)2 _ 2(2)t + C= 12- 32 + C^C = 23
/(i) = 3*2- 2*^+23
60. /"(x) = x^
/'(O) = 6
/(O) = 3
/'W = |.
x^dx = h^ + Ci
/'(O) = 0 + Ci = 6^ Ci = 6
/'(;c) = 3^ + 6
=10'
/(;c) = \{-x^ + 6]dx = —x" + 6x + C2
/(O) = 0 + 0 + Q = 3
fix) = — x" + 6j: + 3
C2 = 3
62. fix) = sinx
/'(O) = 1
/(O) = 6
/'W = I
sin X £& = — cos X + C
/'(O) =-l+Ci = l=>Ci = 2
/'(x) = — cos X + 2
•(^) = J
fix) = I (- COS X + 2)dx = — sinx + 2x+ C2
/(O) = 0 + 0 + C2 = 6 => Q = 6
fix) = -sinx + 2x + 6
Section 4.1 Antiderivatives and Indefinite Integration 453
64. ^ = k^t, 0 < f < 10
dt
Pit)
= ffa'/2A =
rfa3/2 + C
P(0) = 0 + C = 500 => C = 500
PW = t/c + 500 = 600 => A: = 150
Pit) = T{150)f3/2 + 500 = i00r3/2 + 500
P(7) = 100(7)3/2 + 500 = 2352 bacteria
66. Since/" is negative on (-co, 0),/' is decreasing on
(-00, 0). Since/' is positive on (0, co),/' is increasing
on (0, oo)./' has a relative minimum at (0, 0). Since/' is
positive on (-oo, oo),/is increasing on (-oo, oo).
68. /'to = ait) = -32ft/sec2
/'(O) = Wo
/(0) = ^o
fit) = v(r) = I
/'(O) = 0 + C, = vo =» C,
32d?= -32r + C,
/'(f) = -32f + vo
fit) = j(0 = J (-32r + vo) dt= - 16f2 + VqI + Q
/(O) = 0 + 0 + Cj = io =* Q = •5o
/(r) = - 16r- + vof + Jo
70. Vo = 16 ft/sec
ig = 64 ft
(a) j(f) = -16f'+ 16f + 64 = 0
-I6(t^- f-4) = 0
1 ± yi7
Choosing the positive value,
1 + yr?
(b)
~ 2.562 seconds.
v{t) = s'it) = -32r+ 16
<^)
1 + yn
2 / -32^^1 + 16
-16yi7 = -65.970 ft/ sec
72. From Exercise 7 l,/(r) = -4.9?^ + 1600. (Using the
canyon floor as position 0.)
fit) = 0 = -4.9f2 + 1600
4.9^2 = 1600
, 1600
4.9
f = 7326.53 = 18.1:
74. From Exercise 71, f(t) = -4.9F + Vor + 2. If
/(f) = 200 = -4.9f2 + vof + 2,
then
Kr)
-9.8f + Vo = 0
for this f value. Hence, f = Vo/9.8 and we solve
-'■'[ts) ^ "its) ^ ^ = 2o«
(9.8)2 +98 198
-4.9vo= + 9.8vo2 = (9.8)2 198
4.9 vo^ = (9.8)2 198
Vo2 = 3880.8 => Vo = 62.3 m/sec.
454 Chapter 4 Integration
76.
V dv =
-GM
-dy
r
¥^
y
Wheny ■■
= R,v--
= Vo-
y
GM
R
+ C
c
-k
GM
R
2^
GM 1 , GM
IGM
v2 = Vq^ + 2GM|
e-a
78. x(t) = {t- l)(t - 3)2 0 < r < 5
= t^ - 7f2 + 15f - 9
(a) v(r) = x'it) = 3r2 - Ut + 15 = (3f - 5)(r - 3)
ait) = v'(r) = 6f- 14
(b) v(r) > 0 when 0<r<-and3<?<5.
(c) a{t) = 6t - 14 = 0 when / = -.
^«l)-)(l-)-H)-
80. (a) a{t) = cos t
v{t) = J a(t) dt = I cos t dt = sin t + Ci = sin t (since Vq = 0)
/W = I v(r) rff = I sin f rff = -cos t + Cj
/(O) = 3 = -cos(O) + C, = - 1 + C2 => C2 = 4
fit) = - cos r + 4
(b) v(r) = 0 = sin f for f = kir, k = 0,\,2,. . .
82. v(0) = 45 mph = 66 ft/sec
30 mph = 44 ft/ sec
15 mph = 22 ft/sec
ait) = -a
v(l) = -at + 66
■y(f) = -|f2 + 66t (Leti(O) = 0.)
v(?) = 0 after car moves 132 ft.
—at + 66 = 0 whenr
66
a '
■66\ ^ a (66
2\a
+ 661
?)
= 132 when a
33
= 16.5.
ait) = - 16.5
v{r) = - I6.5t + 66
sit) = -8.25t2 + 66f
(a)
- 16.5r + 66 = 44
22
/ 22 \
<,6.5)-^3.33ft
(b)
- 16.5r + 66 = 22
44
t = ^^^^ ^ 2.667
.(^) = 117.33 ft
(c)
#
^
73.33 117.33
feet feet
It takes 1.333 seconds to reduce the speed from 45 mph to
30 mph, 1.333 seconds to reduce the speed from 30 mph
to 15 mph, and 1.333 seconds to reduce the speed from
15 mph to 0 mph. Each time, less distance is needed to
reach the next speed reduction.
Section 4. 1 Antiderivatives and Indefinite Integration 455
/■30 r30
84. No, car 2 will be ahead of car 1. If v,(f) and V2(f) are the respective velocities, then \v2{t)\dt > \v^(t)\dt.
Jo Jo
86. (a) V = 0.6139f3 - 5.525r" + 0.0492f + 65.9881
,^^ ,, , ,,^ 0.6139?* 5.525/^ , 0.0492^2 ^^„„„,
(b) i(f) = v{t)dt = : : — + z + 65.988U
= jvO
4 3 2
(Note: Assume ^(0) = 0 is initial position)
s(6) = 196.1 feet
88. Let the aircrafts be located 10 and 17 miles away from the aiiport, as indicated in the figure.
v^(t) = Z:^ r - 150 Vg = kgt - 250 Airpon
■Sa(') = h^ '- - 150: +10 Sb = ^kg e- - 250t + 17
A -•— B
H h-
(a) When aircraft A lands at time r^ you have
^aUa) = ^a ?a - 150
100
50
^AiO = ^I^A tl - 150r. + 10 = 0
"A^'-Al t'^A 'a
1/50^,
^ 150r^=-10
125f^ = 10
= i°-
'^ " 125-
50
125
625 ,
/t^ = — = 50(— -) = 625 =^ S^it) = 5,W = ^^t-- 150f + 10
Similarly, when aircraft B lands at time tg you have
135
Vgitg) = kg tg " 250 = " 1 1 5 => Ag =
SB{tg)=-kgtl-250tg+ 17 = 0
U—]ti - 250tg = - 17
365
2 *
2Vr '''^
34
'« " 365-
135 ,.,,/365\ 49,275 ^,, ^,, 49,275, .,,„
(b) 20
(c) d = Sgit) - s^{t)
Yes, tf < 3 for f > 0.0505.
20
^
3
/
— ^1
90. True
92. True
456 Chapter 4 Integration
94. False. / has an infinite number of antiderivatives, each
differing by a constant.
96. j^{s{x)Y + [c{x)i\ = 2s{x)s'(x) + 2c{x)c'{x)
= ls{x)c{x) - 2c{x)s{x)
= 0
Thus, \s{x)Y + \c{xJY = k for some constant k. Since,
5(0) = 0 and c(0) = 1, A: = 1.
Therefore,
[six)f + [cixW = 1.
[Note that s{x) = sin x and c{x) = cos x satisfy these
properties.]
Section 4.2 Area
2, ^k(k-2) = 3(1) + 4(2) + 5(3) + 6(4) = 50
*=3
4.^1=1+1 + 1 = 47
^•,4y 3 4 5 60
4
I
1 = 1
6. ^[(/ - 1)2 + (i + 1)3] = (0 + 8) + (1 + 27) + (4 + 64) + (9 + 125) = 238
15 5
10.
l['-(i
16. ;£(2(-3) = 22'-3(15)
i=l i=l
= 2
15(16)
45 = 195
9 "
12. -y
[■-(!-')!
10 10 10
18. 2(r-l) = X''-Il
1=1 /=1 (=1
rio(ii)(2i)"
L 6
"•;sv^
10 = 375
10
1
i=l
10 10
20. 2*^ + 1) = 2^'' + E'
1=1 1=1
102(11)2 , r 10(11)
L-
3080
22. sum seq(x 03 - 2x, x, 1, 15, 1) = 14,160 (71-82)
■g(;3 _ 2i) = (15)'('5 + ^y _ ; 15(15 + 1)
4
(15)2(16)2
15(16) = 14,160
24. 5 = [5 + 5+4 + 2](1) = 16
5 = [4 + 4 + 2 + 0](1) = 10
26. 5 =
2 1
5+2+1+-+-
2 + 1+l + Ul"
3 2 3
55
6
28. 5(8)
1 , „\1
+ 2^. L/l + ^U. /!.zU.(yr.z)l
lV2,V3,,,V5,V6,y7
--/^-^
= 7! 16 + ^ + ^ + ^+ 1 +-V + -^ + ^+ VIH 6.038
4V222 222'
5(8) = (0 + 2)- +
'1 _\1
-a*'Ia*
v^h
7 + 2)7 -5.685
4 14
Section 4.2 Area 457
3...,5,=.(i)./Tf(i).y
-i
-'Di^v-gmi^v-erd
Ifj ^ V24 ^ v^ _^ yi6 _^ 79"
= 0.859
„ ,- r/64\n(« + 1)(2« + 1)1 64,. \2n^ + ?,n~ + n
32. hm — H^ ^ ^ = -r'™ 3
34. hm -r „ =xlim ^ = -(i) = -
6^ -^ 3
36. 2^^V^ = -3E(4/- + 3) =
An(n + 1)
+ 3/j
2/1 + 5
= 5W
5(10) = § = 2.5
5(100) = 2.05
5(1000) = 2.005
5(10,000) = 2.0005
,„ ^APJi- 1) 4^3 .,
n4
+ 1)2 n{n + l)(2n + 1)
]
^ 4r»^ + 2«- + n _ 2n- + 3« + 1]
" /j4 4 6 1
= T-?[3«3 + 6^2 + 3« - 4/j- - 6« - 2]
3«
= Jir[3n' + 2n^ - 3n - 2] = S{n)
5(10) = 1.056
5(100) = 1.006566
5(1000) = 1.00066567
5(10,000) = 1.000066657
40. lim y - - = lim — "V (' = lim
4 /n(w +
nA 2
limfl+iU2
n— *oo Z\ /I/
'>iV/2
42. lim y 1 +- - = lim ^Y (/j + 2/)-
n->oo/e^,\ nj \n/ n->oo /I-",^,
= lim 4rn3 + (4„)(^?(^^ + 4(.)(. + 1)(2„ +0
r 2 4 2 2 1
2 lim 1+2+- + - + - + -^
n->cc L /I 3 /I 3n-J
=21+2+
4\ 26
458 Chapter 4 Integration
2iY 2
44. Urn 2[i +-)[-] = 2\im -^^{n + 2i)
nj \nl n-»oo n^
1 ^
2 lim -7 y (n^ + 6nH + I2ni^ + Si^)
n->oo n^ i^,
2 1im (l+3+- + 4 + - + ^ + 2+- + 4
n->oo \ n n n'- n n''
= 2 lim I 10 +
/I— »co
20
46. (a) y
(b) Ax =
3-1 2
n n
Endpoints:
n n n
■(!)
i<i.iiei<i + 2g)<..-<,+(,-i)g)<i.„g)
(c) Since y = j: is increasing, /(m,) = /(j;,_ ,) on [x,^ ,, jc,].
1/
M:[
iM.--i)(^
i.(.-i)^^
(d)/(M,.)=/(Ac,)on[;c,_„A:,]
(e)
j:
5
10
50
100
^W
3.6
3.8
3.96
3.98
5(«)
4.4
4.2
4.04
4.02
,![-"-«
(f) lim y\\ +0- 1)
limp
n->oo \n
n +
2/«(« + 1)
")]
lim 2
1 + (•
,. [„ 2n + 2 4] ,. r, 21 ,
= hm 2 + = hm 4 =4
n^oo L n n\ n-^oa L n]
^] = lim ^n + P)^^^l
nJ n^oo n\_ \n) 2 J
= lim [2 + 21:^^1=,, [4^21 ^
Section 4.2 Area 459
48. y = 3a; - 4 on [2, 5]. [Note: Ax = = - j
5W = 2/2 +
3A/3
i=lL
3(2 + ^1-4
3A .18 ^3(1^2' -12
, 27/(/j + \)n\ ^ 111, , 1
Area = lim S{n) = 6 +
27 39
SO. y = x^ + 1 on [0, 3]. Note: Ax
3-0 3
*)=i/(?)e)=i,[(!r-B
97 n •J n
^ 27 n(w + l)(2n + 1) , 3 , , ^ 9 2/i^ + 3« + 1
w^ 6 n 2 n'^
Area = lim S{n) = -(2) + 3 = 12
« — *oo 2
1 »-x
SI. y = \ - x~ on [- 1, 1]. Find area of region over the interval [0, 1]. I Note: A.r = -
*.=i/e)(M[.-(i)l
, 1 ^.. , n(>z + l)(2n + 1) 1/9,3,1
= 1 - ^ X(- = 1 —. = 1 - 7 2 + - + —
rv' /e', 6n^ 6\ n n-
1 12
- Area = lim s(n) = 1 - t = -
Area
4
54. >' = Ix: - x3 on [0, 1]. [Note: Ax = = -j
Since y both increases and decreases on [0, 1], T(n) is neither an upper nor lower sum.
= lV-lv-3 = "(" + 1) _ 1 \"Hn + 1)']
1 +
1 1
1
4 4n 4n-
Area = lim T(n) =1-7 = 7
n->oo 4 4
460 Chapter 4 Integration
56. y = )? - ^ oni- 1, 0]. [Note: Aa: = ^^ = -j
.»)=i/(-.4)(^i[(-V-(-i)i
iW^'i^f-'m-^-nrn:'-74;
2 In 3 n "*" 3«3 4 2n 4n2
,-■ / ^ . 5 4 1 7
Area = hm s{n) = 2-- + --j = T:r
58. gCy) =^y,2<y<4. (Note: Ay = ^^^ — - = -
Sin) = 2 ^(2 +
n +
1 /;(« + 1)
2 +
« + 1
Area = lim S{n) = 2+1=3
12 3 4 5
60. f(y) = 4y-yM <y <2. f Note: Ay = = -
^W = E/(i+f)(i
1 "
4„.i- ,.i
= ^3n+^^
n\_ n
n n
+ 1) 1 n{n + l)(2n + 1)
3 +
« + 1 (n + l)(2n + 1)
Area = lim S{n) = 3 + 1
i = ii
3 3
12 3 4 5
62. /;(>■) = y + 1, 1 < >- < 2 I Note: Ay = -
«")=IX'+^)(i)
n /^[ \ n' n'^ n
In +
1 nKn +\Y , 3 «(« + 1)(2» + 1) ^ 3 n{n + 1)
+ ■
. , (« + IP , 1 (n + l)(2n + 1) , 3(n + 1)
n24 2 n^ 2/i
Area = lim Sin) = 2 + 7+l+| = ^
n -trjj 4 2 4
Section 4.2 Area 461
64. fix) =x^ + 4x,0 < X < 4,n = 4
Let c, =
Xi + x.
66. f{x) = sin X, 0 < X < -, n = 4
Let C; =
Aj: = L c, = -, c.
Area=|;/(c,)Ax= ^^r + 4c,](l)
i")^(!-)-(f-»)^(f-*
53
.77 77 Stt
_ Stt _ Vtt
Area « ^-^^'''^ ^ = E (^'" ^^^
= 1
77-/' . 77 377 -577 . 777\
68. /(x)
^2+ 1
on [2, 6].
«
4
8
12
16
20
Approximate area
2.3397
2.3755
2.3824
2.3848
2.3860
70. fix) = cosv/x on [0, 2].
n
4
8
12
16
20
Approximate area
1.1041
1.1053
1.1055
1.1056
1.1056
72. See the Definition of Area. Page 259.
74. fix) =l/x,Q < X <
n
10
20
50
100
200
sin)
10.998
11.519
11.816
11.910
11.956
Sin)
12.598
12.319
12.136
12.070
12.036
Min)
12.040
12.016
12.005
12.002
12.001
(Note: exact answer is 12.)
76.
78. True. (Theorem 4.3)
a. A = 3 square units
462 Chapter 4 Integration
lir
80. (a) d =
(b) sin 0 = -
r
h = rsin 6
A = -bh = -r{r sin 9) = -i^ sin 0
(c) A„ = n\ -r~ sm — = — sin — = -nr-^f
1 , . 27r\ r-n . 2it
-r~ sm — = — - sin —
2 n J 2 n
lir/n
Let.r = 2Tr/n. As « — > oo, x — > 0.
lim A„ = lim Trr- 1 = irr^l)
n-»oo jr->0
82. (a) J]2i = n(n+ 1)
The formula is true for « = 1: 2 = 1(1 + 1) = 2
Assume tliat the formula is true for n = k:
k
^2i = k{k+ 1).
k+l
I
Then we have ^ 2i = ^2/ + 2(/t + 1)
= k{k + 1) + 2(A: + 1)
= {k+ l)(/t + 2)
Which shows that the formula is true for n = A: + 1.
The formula is true for n = 1 because
P(1 + 1P_4_
4 4
Assume that the formula is true for n = k:
e{k + 1)2
k+\ k
Then we have ^ i^ = "^P + {k + 1)
i=l ;=1
k\k + 1)2
4
ik + 1)2
4 ^^
ik 4- 1)2„
+ ()t + 1)3
{k^ + 4(k + 1)]
ik + 2)2
which shows that the formula is true foTn = k+ 1 .
Section 4.3 Riemann Sums and Definite Integrals
2. f(x) =l^,y = 0,x = 0,x=l,c^ = ^
Ax,
i^ a - 1)3 3(2 - 3t + 1
n-" w
n n /;3 r^,-2 _ ^; 4. 1 "1
lim 2 /(c,) Ax, = lim X V T ^^ T^
n-.cc ,^j n->os ,^j V n^ L 1 J
lim ^2(3'-'- 3'' + 0
= lim —
n->oo n
= lim
lim —
1— »oo /Z
n2(„ + 1)2
Jnjn + l)(2n + 1)\ ^ n(n + 1)1
3n* + 6n3 + 3«2 _ 2n^ + 3n^ + n n- + n]
4 2 "^ 2 J
3/2" , n^ „2-| 1-3 1 1 ] 3
Section 4.3 Riemann Sums and Definite Integrals 463
4. >> = JT on [-2, 3]. Note: Ax =
3 - (-2) _ 5
n n
► 0 as n-^oo
2/W^,= 2/(-2.^B = 2U.|)(5) = -,o.iy,
r
-,„.(|)*fil = -,o.f(,.i).^i
j:a:x= lira - + — =-
, n^oo \2 2n/ 2
6. y = Ix'^ on [l, 3]. (Note: Ajc =
3-1 2
2A/2
X3 1 +
n i\n
liVIl
n \n
>0 as «^oo
4 n(n +1) 4 «(n + l)(2n + 1)
« "I :: 1 — ^ ::
= 6+12^^ + 4(^^±il%l^
r
3x'dx = lim
. _^ 12(n + 1) ^ 4{n + 1)(2« + 1)
= 6 + 12 + 8 = 26
8. y = 3x= + 2 on [- 1, 2]. (Note: Ax =
|/Wi., = |/(-i.f)(2
2-(-l)_3
is
3(-l +-1 +2
«/
3|l-^ + ^K2
n n
>0 as «— >oo
3n
18 n(n + 1) ^ 27 n(n + 1)(2« + 1)
3„]
£.'"=
= 15 _ 27(n + 1) _^ 27 (n + Djln + 1)
n 2 «'
+ 2).. = lim [15 - 27^^^^ + ^k±_feM)l
n ->oo L /I 2 n' J
= 15 - 27 + 27 = 15
10. lim V6c,(4- qpAt, = 6x(4-x)2dr
IAI-.0 ,/e, Jo
on the interval [0, 4].
12. lim vfAKv, = \ ^dx
on the interval [1. 3].
Jo
(4 - 2x) (ic
16. f'x-
Jo
dx
18. f ^
(it
r»/4
20. t:
Jo
tan X dx
464 Chapter 4 Integration
Jo
22. {y-2Ydy
24. Rectangle
A = bh = 2(4)(a)
J -a
4dx = ^a
5-
": 24- -S^ Rectangle
f m
pmm — -.
-a I a
28. Triangle
A = ^bh = |(8)(8) = 32
30. Triangle
A = |m = ^(2a)a
Jo
x)dx = 32
A =
x\) dx-- a^
J '4 r4 r4
x^dx = 60,\xdx = 6,\dx = 2.
2 Jz J2
26. Triangle
A=\bh = |(4)(2)
32. Semicircle
A = ^Trr^
A= fy
a,.
r^ - x^dx--
y
u
r
Semicircle
jpy^''
*^
^3^ = 0
r4 r4 ("4
38. (a:3 + 4)a:r = Legate + 4 Lie = 60 + 4(2) = 68
42. (a; \ f{x) dx = \f{x)dx+ \f{x)dx = 4 + (-1) = 3
Jo Jo h
(b)
\f{x)dx= -J
f(x)dx= -(-1) = 1
(c) fix) dx=0
(d)
-5fix) dx= -5\ fix) dx= -5i-l) = 5
36. I 15dx=l5\ dx= 15(2) = 30
/•4 /•4 r4 r4
40. (6 + 2;i: - jc^)^ =6 dx + 21 xdx - x^dx
= 6(2) + 2(6) - 60 = -36
44. (a) j fix) dx= \ f(x)dx- ifix)dx = 0-5 =
(b) fix) dx- \ fix) dx = 5 - i-5) = 10
(c) j 3/(;c) dx = 3j fix) dx = 3(0) = 0
(d) 3fix) dx = 3\ fix)dx = 3(5) = 15
Jo Jo
Section 4.3 Riemann Sums and Definite Integrals 465
46. (a) [f(x) + 2]dx = f{x) dx + 2dx = 4+ 10= H (b) f(x + 2)dx= f(x) dx = 4- (Let // = ;c + 2.)
Jo Jo Jo J-2 Jo
1>
(c) fix) A = 2 fix) dx = 2(4) = 8 (/ even)
(d) fix) dx = 0 if odd)
48. The right endpoint approximation will be less than the
actual area: <
50. The average of Exercise 39 and Exercise 40 consists of a
trapezoidal approximation, and is greater than the exact
area: >
52. fix) = \x\/x is integrable on [— 1, l], but is not contin-
uous on [— 1, 1]. There is discontinuity atx = 0. To see
that
r
M
dx
is integrable, sketch a graph of the region bounded by
fix) = \x\/x and the .t-axis for - 1 < .x < 1. You see that
the integral equals 0.
-i — 1^
H 1-
56. >
1: 2' 3: * 5! 6: 7: 8; *
c. Area = 27.
Jo
60. I X sin ,r dx
54.
1 — I — *-'
b. A = 3 square units
Jo -1^- + J
dx
n
4
8
12
16
20
Lin)
2.8186
2.9985
3.0434
3.0631
3.0740
Min)
3.1784
3.1277
3.1185
3.1152
3.1138
Rin)
3.1361
3.1573
3.1493
3.1425
3.1375
n
4
8
12
16
20
Lin)
7.9224
7.0855
6.8062
6.6662
6.5822
Min)
6.2485
6.2470
7.2460
6.2457
6.2455
Rin)
4.5474
5.3980
5.6812
5.8225
5.9072
62. False
f..v^ ...(/;...-)(/;
64. True
'x dx
66. False
i:
V dx = 6
466 Chapter 4 Integration
68. f(x) = sin;c, [0, 2Tr]
Xq = 0, Xy = ~r, X2 = —, Xj = 1T,X^ = LIT
TT TT '2.TT
TT
TT 2t7
377
-'"6' ^2 -3. -3 3-M 2
i /(c,) .y., = /(f) Ax, + /(f) ^, + /(l^) ^X, + /(f) Ax,
=(i)(f)^(f)te)^(f)(f)^(-)(^)-o-
70. To find /o W dx- use a geometric approach.
H M It 1 1 — »
1 2 3
Thus,
Jo
ldjc= 1(2 - 1) = 1.
Section 4.4 The Fundamental Theorem of Calculus
2. f{x) = cos X
TT
cosxdx = 0
Jo
4. fix) = .tV2 - X
r2
r
,rV2 — X dx is negative, -z
\.v = [3v];
3(7) - 3(2) = 15
|V3„
+ 4) (fv =
--v^ + 4v
■fH-20 -(-6 + 8) = -f
-I
10. (3jc2 + 5x- A)dx
, 5x2
x3 + 4x
2
3 / 45
= 27 + 12
I \ 2
1+2-^
12.|V-90.. = [i.-f,2];_
= 38
1 9\ /I 9
4 2/ V4 2
-i:('-z^)-[^^i:
;=(i-,)-(.-iu-.
Section 4.4 The Fundamental Theorem of Calculus 467
£v'/3rf, = gv^/3]'^^ = l[(yr3)4] _ (yz3)4] = o
^f
^dx= J2\ x''/^dx =
72(2)x'/2l* = [272^1* = 8 - 2v^
20. I (2 - t)Jtdt = f (2f>/2 - r3/2) ^, = r|,3/2 _ 1^/2!^ = [7^(20 - 6r)l^
22. I ^^^ Cix = I I U2/3 - ;c5/3) dx
J-8 2^ 2j_8
¥<»--)=¥
3 3
5 8
^n
80
;24 - 15;c)
' = -^(39) + §(144) = ^5^^
80' ' 80'
80
24. {I - \x-l\)dx= [i + {x-?,)]dx+ [3 - (x - 3)] (&
(6 - jc) cit
.2.
3
1
[-f]
/9
I2"
4)-
f
(24-
4+16-18
9 ^ n
2 ~ 2
26. fV^ - 4;c + 3 1 ir = \\x^- - Ax + 3) dx - \\x^ - Ax + I) dx + \\x^ - Ax + 3) dx *^P'« "P *e integral at the zeros
Jo Jo Ji J3 -": = 1- 3)
J - 2x2 + 3;c
.^3
- Zx~ + 3x
x"
2x- + 3x
= [j - 2 + 3 j - (9 - 18 + 9) + f| - 2 + 3 j + (y - 32 + 12 ) - (9 - 18 + 9)
4 4 4
=T-0+T+T-0=4
3 3 3
28.
30.
■7T/4
de= I de =
0 4
Jo COS' e Jo
r-^n ^ r yn
(2 — csc^ x) dx = \2x + cotxl = (ir
32. £\2. + cos t) dt = [r^ + sin rj^;;^^ = (f + ^ " (t " '
+ 0)-(f+l)=f-l=^
34. P = - I s[nddd =
1T
-]7r/2
•1 -
r
36. A = (1 -;c'»)atc= x-
40. A = (.T + sin x) dx = \-i — cos .v
0 yi'^ 1 ■)
-- cos e I = --(0 - 1) = - - 63.7%
38. A = \dx = -- " = -^ + 1 = ^
Ji-t- L A-Ji 2 2
= — + -) = '""""'"'*
468 Chapter 4 Integration
42. Since y > 0 on [0. 8],
Area= (1 + x^'^)dx =
Jo
44. Since y > 0 on [0, 3],
r
(3x - x^dx
X + -x^l^
4
3 , y?
2 3
= 8 + -(16) = 20
0 4
3^9
0 2-
r
46. I -^dx
■2;c2
^=4.? = 4
2 2
/(c)(3 - 1) = 4
I-
-I
r77-/3 r -|7r/3
48. cos j: (ic = sin j: = V3
J-,r/3 L J-^/3
373
52.
cos
'^- 277
c^±
0.5971
1 f'^
dx =
'2 1"'^
— sinx =
.77 Jo
2
/ ,„v „ cos A
(it/2) - Ojo
77
2
Average value = —
77
2
cos j: = —
77
X - 0.881
54. (a) I fix) dx = Sum of the
areas
= |(3 + 1) + |(1 + 2) + |{2 + 1) + (3)(1)
c = ^1 --- 1.6510
50.
3 - 1
^d!x = 2f(l +x-2)dtc = 2L
26
3
(b) Average value =
i'
fix)dx
8 4
7-1 63
(c) A = 8 + (6)(2) = 20
20 10
Average value = "7" = ~;~
\a.
2. --
12 3 4 5 6 7
12 14 5 6 7
Section 4.4 The Fundamental Theorem of Calculus 469
^f
56, I fix) dx = (area or region B) = \ f{x) dx - fix) dx 58. - 2fix) dx = -ll fix) dx
Jo Jo Jo Jo
-2(-1.5) = 3.0
= 3.5 - (-1.5) = 5.0
= U^
60. Average value = - 1 /(x) tic = -(3.5) = 0.5833
64. P = 5(7f + 30'
(a)
62.
^ojy
2\ J '^
«'-!
" _ 2kR-
0 " 3
t
1
2
3
4
5
6
p
155
157.071
158.660
160
161.180
162.247
1 QC4 150
Average profit = 7(155 + 157.071 + 158.660 + 160 + 161.180 + 162.247) = — -. — == 159.026
6 o
(b) I j 5(7f + 30) dt = ^ si^^'- + 30r
6.5
954.061
J5.0 6
(c) The definite integral yields a better approximation.
66. (a) R = 2.33f^ - 14.67f3 + l.61i^ + 70.67r
(b) 100
= 159.010
Jo
(c) Rit) dt =
2.33f5 14.67;* 3.67f^ 70.67f-
5 4
= 181.957
68. (a) histogram
N
,.
18--
16--
14--
12
123456789
(b) [6 + 7 + 9 + 12 + 15 + 14 + 11 + 7 + 2]60 = (83)60 = 4980 customers
(c) Using a graphing utility, you obtain
Nit) = -0.084175f3 + 0.63492r2 + 0.79052 + 4.10317.
(d)
Jo
(e) Nit)dt'^ 85.162
Jo
The estimated number of customers is (85. 162)(60) = 5110.
(f) Between 3 p.m. and 7 p.m., the number of customers is approximately
Hence, 3017/240 = 12.6 per minute.
( I Nit) dtjieO) =
(50.28)(60) = 3017.
470 Chapter 4 Integration
70. F{x)
= i^ + 2t-2)dt =
B-'
" X
- 2r
_2
= ( J + ;c2 - 2;c) - (4 + 4 - 4)
= ^ + ;c2 _ 2;c - 4
4
F(2)
= 4 + 4-4-4 = 0 Note: F(2) = ifi + It - 2) dt = 0
h
F(5)
= ^ + 25-10-4= 167.25
4
F(8)
= ^ + 64-16-4= 1068
4
72. Fix)
P-2 P ll
= -^dt=- 2t-^dt = \
1 1
2 X^ 4
F{1)
44-
F(5)
=^4=-^=-«-
F(8)
_ 1 1 _ 15
~ 64 4 ^ 64
74. Fix)
■x -pr
= sin 9^9= -cose = -cos j: + cosO = 1 - cos;c
Jo Jo
Fil)
= 1 -cos 2- 1.4161
F(5)
= 1 - cos 5 ^ 0.7163
F(8)
= 1 - cos 8 = 1.1455
76. (a) tif- + 1) dr = (r^ + t) rfr =
Jo Jo
d
1^ 1,
^ 1 1 x^
= -.X* + -x^ = — (;c2 + 2)
0 4 2 4^ ■*
(b)
cic
1 4_u 1 2
4 2
x^ + ;c = ;c(jc2 + 1)
78. (a)
P..=
2
_^/2
3
= f.3/2_|.|,3/2_8)
I sec r tan t (it = s
Jv3 L
80. (a) I sec r tan t (it = I sec t
V3
= sec X — 2
ir/3
(b)f 1x3/^
16
= rl/2 =
(b) —[sec X - 2] = sec X tan X
82. F(x)
rff
F'(x) =
84. Fix) = I Vtrft
F'(x) = ifx
Jo
86. F(x) = sec' t (if
Jo
F'(x) = sec'x
x2+ 1
Section 4.4 The Fundamental Theorem of Calculus 471
n,) = £,>.,4g>o
F\x) = 0
Alternate solution
F{x) = \ t^dt
= t^dt+ \ t^dt
= -) fidt+ \ ^
Jo Jo
dt
F'(x) = -(-xn-l) + {x^) = 0
90. Fix
.) = f r3.. = [^l = [-^]; = ^ 4^ F'W = 2.-
Alternate solution: F'(.r) = (x^)~\2x) = 2x'
= I sin d-c
Jo
91. F{x) = I sin e-dd
F'ix) = sin (x2)2 (2x) = Zrsinjc"
94. (a)
X
1
2
3
4
5
6
7
8
9
10
gix)
1
2
0
-2
-4
-6
-3
0
3
e
(c) Minimum of ^ at (6, - 6).
(d) Minimum at (10, 6). Relative maximum at (2, 2).
12
(e) On [6, 10] g increases at a rate of — = 3.
(f) Zeros of g: x = i,x = &.
96. (a) g{t) = 4 - ^
(b)
lim g(t) = 4
r->oo
Horizontal asymptote: v = 4
(b) A{x)
r
4-1).
4>
4r + - =
t]\
4.t + - - 8
4x- - 8.T + 4 4(.t - 1)-
lim A(x) = lim4.r + --8=oo + 0-8 = oo
X— >o<: x->o<: \ .t /
The graph of A(x) does not have a horizontal asymptote.
98. True
100. Let F{t) be an artiderivative of/(r). Then,
d^
dx
J-i-(-t)
/(f) dt
uW
/(f) dt
F{t)
■iti
4-c)
Rv(.r)) - F(,u{x))
u(x)
= ^f(v(.t)) - F{u{x))
= F'(v(.t))v'(x) - F'(M(:t))«'(^)
= f(v(,x))v\x) - f{u{x))u\x).
472 Chapter 4 Integration
102. G{x)
Jo I Jo
f(t)dt
(a) G(0)
Jo L Jo
ds
fit) dt
ds = 0
(c) G"(x) = X ■ fix) + \fit)dt
Jo
(d) G"(0) = 0-/(0)+ \ fit)dt = Q
Jo
104. xit) = it- Dit - 3)2 = r' - 7^2 + 15r - 9
x'(r) = 3f- - 14f + 15
Using a graphing utility,
(b) Let Fis) = J /(f) dt.
Jo
G(x) = F(5) ds
Jo
Total distance
Jo
\dt ~ 27.37 units
Section 4.5 Integration by Substitution
\figix))g'ix)dx u = gix) du = g'ix)dx
2. |xV;c3 + \dx
4. sec 2;c tan 2;c ^
^ + 1
2x
6. I^^dx
2d:x
COS j: dx
8. [u^
9)3(2;c) dx = ^ ^ ' + C
Check: ^
ax
(x^ - 9)"
4
+ C
4(^2 - 9)3
(2x) = ^2 - mix)
../,
10. I (1 - 2x2)'/3(-4x) <fe = ^(1 - 2x2)''/3 + c
Check: -^
^(1 - 2x2)4/3 + C
G'(x) = Fix) =xlfit)dt
Jo
G'(0) = o( /Wdr = 0
Jo
= I • |(1 - 2x2)i/3(-4x) = (1 - 2x2)i/3(-4x)
../.
x^U' + 5)''d:t = ||"(x3 + 5)4(3x2) dx = l^^^y^ + C = ^^^-Js^ "^ ^
Check:
^[(xL+ili , J ^ 5(x3 + 5[
dxl 15 J 15
3 5
5)^(3x2) _ ^
x3 + 5)V
14. |x(4x2 + 3)3 ^ = M (4x2 + 3)3(8j,) ^ = i
(4x2 + 3)4-
.,c = (^^i±^ + c
32
Check:
otc
(4x2 + 3)4
32
+ C
4(4x2 + 3)3(8;t)
32
= x(4x2 + 3)3
Section 4.5 Integration by Substitution 473
ft'VFT5dt = ^f{
16. I t^VFTldt = tI (r» + 5)'/2(4r3) dt = \^^^jp^ '^^^h''^ ^^'''^ "^ ^
Check: 4
7(r* + 5)3/2 + c
6
1 1
6 ' 2
(f* + 5)i/2(4f3) = (f + 5)'''2(t3)
S. mV«^ + 2 ^" = T ("^
18. I «V«3 + 2du = ~\{u^ + 2)'/2(3«2) rftt = ^^ ^r^' +C= ^ ^ ' + C
+ c] = I • |(«3 + 2)'/2(3«=) = («3 + 2)'/2(«=)
20.
'^7)5^ = ^J(l+^")-W)^=4(l+^)"' + C = ^^j
+ c
Check:
dx
x^
(1 + x^P
2^- /(T6^'^ = 4/('^ - -^)-=(-3-)^v = -|[^^^^]
+ C
1
3(16 -x3)
+ C
Check:
dx
^ +c]=|(-l)(16-x3)-2(3x2)= •»
3(16 - x^)
(16 - x^r-
24.
/^-i/'
x^ 1 r 1(1+ x^)'/- /I + v^
Check:
dx
ym?
+ c
4 1/2
-•-(1 +.v4)-'/2(4.v3) = --=^==
2 2 VTT^
26.
/['-i5V]-=/('-H-=f-l(9-=?-^^--
r'- 1
9x
+ C
Check:
<&
^-r' + C
, 1 _. , 1
Check: ^[V^ + C] = ^
30. y-^-T^'it = I (i'/2 + 2f3/2) A = |f3/2 + |rV2 + c = ^^^''-(5 + 6f) + C
Check: 4
2 4
£f3/2 + 1^/2 + c
;l/2 + 2r3/2 =
f + 2f2
Check:
¥*i?
474 Chapter 4 Integration
34. j 2773^(8 - f/^) dy = iTrU^y - f'^) dy = ItUy^ - ^fA + C = ^(14 - y^/^) + C
Check:
dy
477 -r.
14 - y3/2) + c
d_
dy
277(4/ -^7/2 I + C
= \6TTy - l-nfl'^ = (277)')(8 - y^/z)
1. V = , dx
= y|(l + y?)-^l\Zx^) dx
_ ior(i+^3)i^i
1 1/2 J
20
= ^yrT^ + c
40. (a)
M^'^SM
-I--
-3-
M^?W
38. j;
X- 4
--dx
J Vx^ - 8a: + r
{x^ -8x+ l)'/2
1/2
+ C
= V';c2 - 8;c + 1 + C
(b) V- = ^ cos x^, (0, 1)
y = I ;i: cos x^ dx = —\ cos(j:2)2x dx
1
sin (x^) + C
(0, 1): 1 = - sm(0) + C => C = 1
y = - sinU^) + 1
j 4^:^ sin a:^ ^ = I :
42. 4^3 sin;i^^ = sin;d(4x3) dx = -cosx^ + C
I cos 6x dx = -\ {<
1 .
44. I cos 6xdx = -\ (cos 6j:)(6) dx = — sin 6;t + C
5. J X sin x^ ctr = - I (
46. I X sin x^ ctr = - 1 (sin x^){2x) dx = -- cos x^ + C
/^
-/'
48. sec(l - x) tan(l - x) dx = - \ [sec(l - jc) tan(l - x)]{- I) dx = -sec(l - x) + C
»■/■
50. I Vtanlc sec^ ;c dt = ™^^ h C = -(tanx^/^ + C
52. ^'V ^ = - {cosx)-\-smx)dx = - ^ + C = t r- + C = -scc^a: + C
Jcos^x J -2 2cos2x 2
54.Jcsc2(f)^ = 2jcsc^
;c\/l
2/V2
dx = -2 cot - + C
(f)
-I""
56. /(j:) = I 77 sec 77X tan nxdx = sec 77X + C
Since /(1/3) = 1 = sec(77/3) + C, C = - 1. Thus
f(x) = sec 77X - 1.
Section 4.5 Integration by Substitution 475
5%. u = 2x+ \,x = -(u- 1), dx = ]rdu
I xjlx + 1 dx = \hu - \)^]-du
= \\^{u'''-u''')du
„3/2
= 3^(2r + l)V2[3(2x + 1) - 5] + C
2
30
(2;t + \y'\6x - 2) + C
60. M = 2— X, x = 2-M, iit= —du
\{x+ \)J2 - xdx= - (3 -m)v^
= - (3u ' - u
du
3/2
)rftt
= -(2tt3/2-|„5/2)+c
2m3/2
{5 - m) + C
= -j(2-xm5-(2-x)] + C
= -|(2 - ;c)3/2(;c + 3) + C
= ^{2x + l)3/2(3;c - 1) + C
62. Let u = X + 4, X = u - 4, du = dx.
= |(2«'/'-7M-'/2)d«
= |m3/2 - 14„l/2 + C
: jm'/2(2« - 21) + C
64. u = r - 4, r = M + 4, rff = rfu
I t^t- 4 dt= (« + 4)m'/3 dtt
/•
= (« + 4)m
,4/3 +4„l/3)d„
= ^«'/3 + 3„4/3 + C
3«^/3
iu + 1) + C
= ->A + 4[2(-T + 4) - 21] + C
= =jit- 4)''/3[(r - 4) + 7] + C
;-VFT4(2x - 13) + c
= ^r - 4)^/3(f + 3) + C
66. Let « = x^ + 8, rfM = 3^^ dx.
r
c
x2(x3 + 8)2 ate = ^ I (x^ + 8)2(3^2) die
3 3 J--
= ^[(64 + 8)3 - (-8 + 8)3] = 41,472
68. Let M = 1 - .r^, du = -Ixdx.
f x^V^^dx = -||" (1 - .t2)'/2(-2v) cfe = r-|(l - .X'V
:>-r;
70. Let M = 1 + 2x-, c/m = 4x dx.
C X ^^1 p
Jo v-m^ ' 4jo
(1 + lx-)-'/-(4.r)d:r = ^71 + li
[i-
^-i=l
476 Chapter 4 Integration
72. Let « = 4 + .x\ du = 2xdx.
xi/4 + x'^dx = ^\
Jo ^Jo
xi/4Tl(^ dx = i-\ (4 + x^y/^{2x) dx=\^{4 + x^Y'^
= ^(S''/^ - 4*/3) = 6 - ^^ - 3.619
0 5 2
74. Let H = 2x - L rfu = 2 A, x = -(« + 1).
When ;c = 1, M = L When x = 5,u = 9.
Ji 72F^M J, v4 2 4ji
^[f^(27) + 2(3)) - (I + 2
4LV3
26
3
,6. £;v . cos,)* = [f . si„,]:;; . (f . ,) - (g . f ) . f
'tt^ , 73\ _ 517^ _^ 2-73
78. u = x + 2, x = u — 2, dx = du
Whenx = -2, m = 0. When a: = 6, m = 8.
Area
( ;C-3/^rT2 die = I (m - 2)2 34rfM = ( (m'/3 _ 4„4/3 + 4„l/3) J„
J-2 Jo Jo
3 ,0/3 _ il„7/3 + 3„4/3
4752
35
Jo
80. A = (sin j: + cos 2x) dx =
- cos j: + - sin 2x I =2
•r
82. Let « = 2x, dM = 2 iic.
rnM J p/4 r J "1^/4 J
Area = esc 2Ar cot 2x dr = - esc 2jccot 2jr(2) dx = -— esc 2;c =-
J7r/12 2J^/i2 L 2 J,r/12 2
Jo
84. x^^x + 2(ic = 7.581
^/:
86. xVx - 1 otc = 67.505
90. I sin X cos X ate = I (sin x)' (cos x ate) = — r— + Q
sin X cos X otc = I
sinxcosxdr = ~
sinxcosxiic= - (cosx)'(— sinxatc) =
+ C,
(1 - sin^x)
^sin^_l
^ (-2 2 2
J-ir/2
si
0
88. sin2xatt = LO
.^
They differ by a constant: €2= C, +
1
Section 4.5 Integration by Substitution 477
92. f{x) = sin^ X cos x is even.
■w/Z
rnn
.
sin^
^ccos j:
dx
= sin^ j:(cos x) dx
-7r/2
Jo
fsin^
L 3
_2
3
-0
94. /(a:) = sin j: cos x is odd.
rir/2
sin X cos j: A: = 0
-ir/2
J-%
rir/4
96. (a) s
J-ir/4
sin X dtc = 0 since sin x is symmetric to the origin.
(b)
rn/4 r-w/i r -|Tr/4
COS xdx = 2] cos j: A = 2 sin X = v^ since cos jr is symmetric to the y-axis.
J-T/i Jo I Jo
r-^/2 r-r/2 r -1^/2
(c) I cos xdx = 2\ cos xatc=2sinjc =2
J-V2 Jo L Jo
p/2
(d) sin X cos .r tic = 0 since sin(— j:) cos(— j:) = — sin j: cos x and hence, is symmetric to the origin.
J-Tr/2
98. j (sin 3x + cos 3x) dx = I sin 3j: lic + cos 3xdx = 0 + 21 cos 3j; tic =
J-w J-TT J-TT Jo L
cos 3x dx = I — sin 3x
= 0
'/'
100. If« = 5 - x^,thendu = -2x&and \x{5 - x^f dx = "T (5 - x2)3(-2x) A = --|«3(i„.
^1'
= 4P
102. ^ = /t(100 - r)^
f
Q{t) = I A:(100 - tf dt = --(100 - f)' + C
(2(100) = C = 0
<2(r) = -jdOO - tf
Q(0) = -j(100)3 = 2,000,000 => /t = -6
Thus, Q{t) = 2(100 - tf. When t = 50, 2(50) =
$250,000.
106. (a) 70
104. R = 3.121 + 2.399 sin(0.524f + 1.377)
(a) ^
Relative minimum: (6.4, 0.7) or June
Relative maximum: (0.4, 5.5) or January
(b) Volume
Jo
if" 1
I R{i)dt-
(b) I R{{) dt = 37.47 inches
(c) 7 1 y?(f) ^f = T-(13) = 4.33 inches
1272 (5 thousand of gallons)
Maximum flow: /?== 61.713 at r = 9.36.
[(18.861, 61.178) is a relative maximum.]
478 Chapter 4 Integration
108. (a) "
(b) g is normegative because the graph of/ is positive at the
beginning, and generally has more positive sections than
negative ones.
(c) The points on g that correspond to the extrema of/ are
points of inflection of g.
(e) 4
^^-^
The graph of h is that of g shifted 2 units downward.
g{t) = I /U) dx
Jo
I f(x)dx+ I f
Jo Jtt/I
fix) dx+ \ f(x)dx = 2 + hit).
(d) No, some zeros of/, like x = tt/2, do not correspond to
an extrema of g. The graph of g continues to increase
after x = ir/l because /remains above the j:-axis.
110. False
Lix^ + l)2(fa = 11 (;c2 + l)(2;c) dx = ^ix^ + l)^ + C
112,
True
•*
sin j:
Ja
dx =
114.
False
i
— cos x\ = - cos b + cos a = - cos(fe + 2
tt) + cos a = I
sin j: lie
I sLn^ 2x cos 2xdx = -\ (sin
1 I 1 (sin2x)' 1
:cos2xdx = -\ (sin 2x)2(2 cos 2x) dx = - - — r—'- + C = - sin^ 2x + C
21 I i o
116. Because /is odd, /(-At) = -/(jc). Then
ra ro fa
fix) dx = fix) dx + fix) dx
J-a J-a Jo
= -( fix)dx+ \fix)dx.
Jo Jo
Let x = —u,dx= — rfM in the first integral.
When X = Q,u = 0. When x = -a,u = a.
I fix)dx= - \fi-u)i'du) + \fix)dx
J~a Jo Jo
= -\ fiu)du+ \fix)dx==0
Jo Jo
Section 4.6 Numerical Integration 479
Section 4.6 Numerical Integration
2. Exact:
f(f
+ 1 Uc =
x^
+ X
1.1667
Simpson's:
4. Exact:
Trapezoidal:
Simpson's:
6. Exact:
i.4|M:.i).2(M)!,,),4(l«.i).g.i
75
= £-1.1719
64
= 1.1667
= 0.5000
l+4|i)= + 2(i)" + 4(f + i].0.5004
12.0000
Trapezoidal: iG dx = ^[0 + 2 + 2^/2 + 23/3 + 23/4 + 2^5 + 2^6 + 2^7 + 2] « 11.7296
Jo 2
Simpson's: .^ otc == :^[0 + 4 + 2^2 + 4^ + 2i/4 + 4^5 + 2i/6 + 4i/l + 2] « 11.8632
Jo 3
8. Exact:
(4 - jc2) dx -■
Trapezoidal: (4 - x^)dx ^ ^3 + 2 4
Simpson's: I (4 - x^) a[r = - 3 + 4(4
1 i
2 ^
"3
+ 2(0) + 2
-0.6667
4
4, + 0 + 4(4-f
iJ]
5 I - -0,
5 = -0.7500
6667
10. Exact:
I
x^x^ + \dx = -
(.^ + 1)3/2
' = i(53/2 - 1) ^ 3.393
0 3
Trapezoidal:
+ 1 A - Ifo + 2(^)^(1/2)- + 1 + 2(i)yFTT + 2(|)v(3/2)^ + 1 + 27FTTJ -
3.457
xjx- + 1 A = ^1 0
Simpson's: j xV^lH'aLc = | 0 + 4[ijV(l/2)2 + 1 + 2(l)VPin' + 4f|jV(3/2)- + 1 + 2v'2^"+T j - 3.39
12. Trapezoidal: J- dx = 7 1 + 21 , ^ =) + 2(^=1=
jov/m? 4 \vi + (1/2)3/ \jY+
Simpson's: , etc = - 1 -I- 4 , =
JoTrn? 6L vvi + (1/2)
vTTTs
1
+ 2'
+ 4
1
1
Vl + (3/2)3/ 3
1
ym3/ \vi + (3/2)3/ 3 J
1.397
= 1.405
Graphing utility: 1.402
480 Chapter 4 Integration
14.
16.
Trapezoidal:
Simpson's:
Graphing utility
Trapezoidal:
Simpson's:
^xsmxdx = —
n/2 16
"tt
Jx sin xdx'=^ -—
.12 24
: 1.458
V
V
77/4
'^nw, f^-i^A^-, f^-C^AA.-, f^-PA^r.
«= 1.430
1.458
0 ^
« 0.271
tanCv2^ Wv V^V4
V 4 y \ 2 / V 4 / \\J A)
tan 0 . 4 tan(^)% 2 tanf^^)^ . 4 tanf^^//^)^ . tanf ^V
V 4 / V 2 ; V 4 y W 4/.
0
12
■ Q.l'il
Graphing utility: 0.256
J-ir/2
Vl + zos^xdx'- -^v/^ + 2Vl + cos^dr/S) + 2Vl + cos2(Tr/4) + 2Vl + cos2(37r/8) + l] = 1.910
0 16
J-7r/2
Vl + cos2A:dx«:^v/2 +4Vl + cos2(7r/8) +2^1 + cos2(tt/4) + 4V1 + cos2(37r/8) + l] = 1.910
0 24
Graphing utility: 1.910
,„ ^ .J, r sinj: , ttF, 2sm(7r/4) 2 sin(7r/2) 2 sm(37r/4) 1 , „.,^
20. Trapezoidal: dx « - 1 + 7;^ + 7^^-^ + — J^ ' + 0 =» 1.836
f ^gJM ^ ^ 4 1 ^ iji'Ml) ^ 2sin(^ ^ 4 sin(3,r/4) ^ 1 ^
Jo ^ I2L 7^/4 7r/2 3Tr/4 J
Simpson's:
Graphing utility: 1.852
22. Trapezoidal: Linear polynomials
Simpson's: Quadratic polynomials
24. f(x) =
fix) =
/'W =
f"U) =
. f'Kx) =
X
+ 1
-1
{x
+ 1)^
2
(x
+ 1)3
-6
ix
+ 1)"
24
(x + ly
(a) Trapezoidal: Error < ,.,-^. (2) = — = 0.01 since
1 2(4 } 96
f"{x) is maximum in [0, 1] when x = 0.
(h) Simpson's: Error < ^^^^(24) = ^^ » 0.0005
since /**"(x) is maximum in [0, 1] when x = 0.
Section 4.6 Numerical Integration 481
26. ru
in [0, 1].
(1 + xY
(a) |/'tx)| is maximum whenx = 0 and |/"(0)| = 2.
12/1
Trapezoidal: Error < 7:^3(2) < 0.00001, n- > 16,666.67, n > 129.10; let n = 130.
f'\x)
24
in [0, 1]
(1 + xf
(b) \f'^*\x)\ is maximum when a: = 0 and |/''"(0)| = 24.
Simpson's: Error < ^(24) < 0.00001. n-* > 13,333.33, « > 10.75; let n= 12. (In Simpson's Rule /i must be even.)
28. fix) = (x + 1)2/3
(a) /'W =
9{x + l)'*/3
in [0, 2].
l/'lt) I is maximum when x = Q and |/'tO) |
Trapezoidal: Error < Ty^l^l < 0.00001, w^ > 14,814.81, n > 121.72; let n = 122.
(b) /WW = -
56
81U + l)i»/3
. [0, 2]
\f''\x)\ is maximum when .r = 0 and [/("'(O)! = — .
ol
Simpson's: Error < t^^(^] < 0.00001, /i"* > 12,290.81, n > 10.53; let « = 12. (In Simpson's Rule n must
, . 1 0O/7 \ 8 1 /
be even.)
30. fix) = sin(.t^)
(a) fix) = 2[-2x= sinU-) + cosix^)] in [0. 1].
\f"ix)\ is maximum when.r = 1 and |/"(1)| = 2.2853.
Trapezoidal: Error < ^' ~ , (2.2853) < 0.00001, n- > 19,044.17, n > 138.00; let/! = 139.
' (b) /W(x) = (16x^ - 12) sin(x2) - 48x^ cos(a^) in [0, 1]
[/(■"(x)! is maximum when x == 0.852 and |/<'"(0.852)| == 28.4285.
(1 - 0)5
Simpson's: Error <
180«''
-(28.4285) < 0.00001, «■» > 15,793.61, « > 11.21;let/i= 12.
32. The program will vary depending upon the computer or programmable calculator that you use.
34. fix) = Vl - .r= on [0, 1].
n
Un)
Min)
Rin)
7t")
Sin)
4
0.8739
0.7960
0.6239
0.7489
0.7709
8
0.8350
0.7892
0.7100
0.7725
0.7803
10
0.8261
0.7881
0.7261
0.7761
0.7818
12
0.8200
0.7875
0.7367
0.7783
0.7826
16
0.8121
0.7867
0.7496
0.7808
0.7836
20
0.8071
0.7864
0.7571
0.7821
0.7841
482 Chapter 4 Integration
36. /W = ^ on [1,2].
n
L{n)
M{n)
R{n)
Tin)
Sin)
4
Q.IQIQ
0.6597
0.6103
0.6586
0.6593
8
0.6833
0.6594
0.6350
0.6592
0.6593
10
0.6786
0.6594
0.6399
0.6592
0.6593
12
0.6754
0.6594
0.6431
0.6593
0.6593
16
0.6714
0.6594
0.6472
0.6593
0.6593
20
0.6690
0.6593
0.6496
0.6593
0.6593
38. Simpson's Rule; « = 8
l-lsm^dde^' ^ V ^ ~ f ^'"' 0 + 4^]
6
^ 17.476
sin2 0 + 4.^/l -|sin2:^ + 2
3 16
>yrri^+...+^
1 2 . , TT
1 — r sm'^ —
3 2
40. (a) Trapezoidal:
I fix)dx^ Wf'^-^^ "^ ^'"^-^^^ "^ ^^"^-^^^ "^ ^^^-"^^^ "^ ^^^-^"^^ "^ ^^''■^^^ "*" ^(^-^^ "^ ^(^-^^^ "^ ^•^''■] " ^^-^^^
Simpson's:
2
/:
fix) dx - ^[4.32 + 4(4.36) + 2(4.58) + 4(5.79) + 2(6.14) + 4(7.25) + 2(7.64) + 4(8.08) + 8.14] = 12.592
(b) Using a graphing utility,
V = - 1.3727x3 + 4.0092x2 - 0.6202x + 4.2844
Integrating,
ydx'='n.
Jo
53
42. Simpson's Rule: n = 6
77 = 4
Jo 1 + ^-
4
3(6)
4 2 4 2 4 1"
1 + . . .. ,,x. + . . .^,... + ■ . ..,„..+ . . ...,..+ . . . +■
1 + (1/6)2 1 + (2/6)2 1 + (3/6)2 1 + (4/6)2 j + (5/5)2 2J
^ 3.14159
120
44. Area = ^7J^[75 + 2(81) + 2(84) + 2(76) + 2(67) + 2(68) + 2(69) + 2(72) + 2(68) + 2(56) + 2(42) + 2(23) + 0]
= 7435 sq m
46. The quadratic polynomial
p(^^ ^ Jl^LJ^imhL I (^ - x,)ix - x^) I jx - x,)ix - X,)
(x, - X2)(Xi - X3) ' (X2 - Xi)(x2 - X3) 2 (jj^ - x^)iXi - Xj) ^
passes through the three points.
Review Exercises for Chapter 4 483
Review Exercises for Chapter 4
2.
4. M = 3x
du = 3,dx
2
/j|-f/
{3jc)-'/3(3)d:t=(3;c)2/3 + C
6- [^^ §~~^^ = f U - 2 + x-2) ^
-j:^ - 2x - - + C
2 j:
10. fix) = 6U - 1)
/'W = 6(;c - 1) dx = 3{x - 1)2 + C,
Since the slope of the tangent line at (2, 1) is 3, it follows
that/'(2) = 3 + Ci = 3 when Ci = 0.
fix) = 3(.t - 1)2
fix) = 3U - \)'dx = (x - 1)3 + Q
/(2) = 1 + C2 = 1 when Cj = 0.
/W = U - 1)3
../,
8. I (5 cos x - 2 sec^ x)dx = 5s\nx-2tanx + C
12. 45 mph = 66 ft/sec
30 mph = 44 ft/sec
a(f) = -a
at + 66 since v(0) = 66 ft/sec.
-t~ + 66r since s{0} = 0.
v(r)
Solving the system
v(f) = -at + 66 = 44
s(t) = -^- + 66r = 264
we obtain t = 24/5 and a = 55/12. We now solve
-(55/12)f + 66 = 0 and get r = 72/5. Thus,
72
■mm.eeiii
" 475.2 ft.
Stopping distance from 30 mph to rest is
475.2 - 264 = 211.2 ft.
14. a(t} = -9.8m/sec2
v(f) = -9.8t + Vq = -9.8f + 40
s{t) = -4.9?2 + 40t (5(0) = 0)
40
(a) v(t) = -9.8r + 40 = 0 when f = ^ ^ 4.08 sec.
(b) i(4.08) = 81.63 m
(c) v(t) = -9.8f + 40 = 20 when t
(d) i(2.04)«= 61.2 m
20
9.8
2.04 sec.
484 Chapter 4 Integration
16. Xj = 2, j:2 ~ — 1, ^3 = 5, ;c4 = 3, X5 = 7
mjinf-h- l+5 + 3 + 7)--!|
37_
5 ■ 3 ■ 7 210
(c) 2(2^,- - ^,') = [2(2) - (2)2] + [2(- 1) - (- 1)2] + [2(5) - (5)^] + [2(3) - (3)^] + [2(7) - (7)^] =
1 = 1
(d) 2(x, - x,_,) = (-1 - 2) + [5 - (- 1)] + (3 - 5) + (7 - 3) = 5
-56
18. y = 9 - -x\ Ax = 1, n = 4
5(4) = 1
9 - i(4)) + (9 - ^(9)) + (9 - ^(16)) + 9 - |(25)]
= 22.5
5(4) = 1
== 14.5
9-^(9)1 +
9 - ^(16)) + (9 - |(25)) + (9 - 9)]
2
20. y = X- + 3, Ax = - right endpoints
n
Area = lim y* f{ci) Ax
= lim ^X Ft + 3
"4 w(«+ 1)(2«+ 1) , , 1
V 1 — ~^H
r^(n+l)(2n+l)^ 1 8^^
_3 w^ J 3
= lim -
n— >oo n
— lim
26
3
1 2
22. >> = -X?, Ax = -
4 w
Area = lim V /(ci) Ax
,. 1 -^Tn 24i 24/2 s/n
lim — y 8 + — + — ^ + — r
n->oc2n-^,L n « n J
n->oo n j^i L " « « J
= 1- if I 3 w(« + 1) ^ 3 n{n + l)(2n + 1) , 1 w2(n + l)^"!
n-^co nL«2 «2 6 n'4j
= 4 + 6 + 4+1 = 15
12 3 4
Review Exercises for Chapter 4 485
- <». ^ ^ m - <m - «(f )(!) ^ "(t)© = if <■ — ' = ^
J = m(0)
(!)
-i!)(!) -(!)(!-»
4JU/ 16'
(1 + 2 + 3) =
8
0\ wfc^(« + 1)
2n
•« = lao = !"(!)© = -er % - """- ^ "-^ - ""--^^'
«^
2«
(c) Area = lim ^?^%1I) = u„, ^^^^n - 1) ^ 1 , ^ 1 ^^^^^^^ ^ ^ (base) (height)
n->oo 2n n-»oo 2n 2 2 2
f ri 1* 1
(d) I mxdx = \ -nvxP- = -wi^
26. lim V3c((9 - c/^) ^xi = 3;c(9 - x^) dx
28.
r.
Vl6 -jc^atc = x'^^^)-
Stt (semicircle)
30. (a) J /(;c)^ = j /(^)dx + \ f{x) dx = 4 + (-1) = 3
(b) J/(;c)(&= -J/(x)d:x= -(-1)= 1
(c) \f{x)dx = Q
(d) - \Qf(x) dx= -\o\ f(x)dx= - 10(- 1) = 10
„ Pl2 , ri2x-21' r-67 -6 , 16 „,
34.J_V + 2)..= [f.2r];_=f
£
36. I {x!^ + Ix- - 5) dx =
-{
L
32 16
+ 10
52
15
^«-f(i4)'^ = />-^-^-')^ = [-^2^
- /
' 1 1\
/ , 1\
1
-1 + -
I
^ 2 8^
V 2/
8
rw/4 r -1^4
I. sec^ r rf/ = tan n = 1 - (- 1) = 2
J-ir/4 L J-7r/4
486 Chapter 4 Integration
i
42. {x + A)dx =
V
+ Ax
10
-2-1 12 3 4 5
44.
/>
x^ + X + 2) dx =
x^ x^
■I— )-(i4-)
= 12 7 ^9
3 6 2
46. V^(
Jo
\-x)dx^ (;c'/2 - ^3/2) dx
2^/2 _
.3
2 T
5 Jo
2 2
3 5 "
4
" 15
48. Area
rV3
Jo
= tanjc
sec^ j: etc
= V3
;c = 3/2
50. :r^ \ x'dx =
x^ = 2
y
= 2
52. F'ix) = -
x^
54. F'(x) = csc2;c
56. (^ + ;^) dx^ I (jc2 + 2 + ;c-2) die
j;3 1
3 X
Review Exercises for Chapter 4 487
58. u = jc' + 2,, du = 2>x^ dx
jx^VJ^TI^ = \U^ + 3)'^' 3j;2alr = |(x3 + 3)3/2 + c
60. M = jr^ + 6j: - 5, (ftt = (2x + 6) ^
\{x4V-5f = ij(.2?6x-5P 'i^ = ^U^ + 6.-5)-' + C= 2(;,2 + ^ _ 5) + C
62. .r sin 3x^ dx = -\ (sin 3Ar2)(6.t) iir = -7 cos 3x- + C
64. M^^£it= (sin;c)-'/2j,os^^ = 2(sinjc)'/2 + c = 27ii^ + C
J Vsinjc J
66. I sec 2.r tan 2r ^ = - I (sec 2x tan 2x)(2) (ic = - sec Ir + C
68. I cor' a CSC- ada = - (cot a)''(-csc- a) da = --cot^a + C
f ;cV+ l)3A = ^f I
Jo -^Jo
70. I ;cV + 1)3 A = 11 (x3 + l)3(3.r:)d:x = ^
Jo 12
(x3+ 1)^1 =^16-1) ^
72.
f ^::#^'^ = C^-^ ~ 8)-./2(2x)^ = [f(^ - 8)-]; = 1(277 - 1)
74. m = j:+1,x = m— l,cir = rf«
When x= - I, m = 0. When .r = 0, m = 1.
76.
27r| xVa: + 1 dx = 27r (« - D^Judu
= 2tt\ (tt5/2 _ 2„3/2 + „l/2) ^„ = 27Tf|«'/2 _ l„5/2 + ^ 3/2I' = HE
Jq L' 5 3 Jo 1U5
r-n/4
sin 2r
J--7r/4
It otr = 0 since sin 2a: is an odd function.
78. « = 1 — x,x = 1 — u, dx = —du
When .X = a, M = 1 -a. When .v = fc, « = 1 - fc.
^0= I
Ja
''^155-.v3(l-.)3/2^.= l'55'—
32
32
Jl-n
(1 -m)3m3/2^„
1155
32
(„9/2 _ 3„7/2 + 3„5/2 _ „3/2) ^„ = 1122 ^„112 _ r„9/2 + £„,,,, _ £^^,
Jl-a 32 [11 3 7 5 Jl-n
= -^ 7777(105"' - 385m= + 495« - 231) = -r^(105u3 - 385m- -i- 495m - 231)
32 L1155 Jl-n L lo Ji-o
(a) P,
(b)
mV2
16
(lOStt' - 385m= + 495m - 231)
"10.75
- 0.025 =
0. 0.25
r,,5/2 no
^oj 1 = Tr(105M3 - 385m- + 495m - 231) = 0.736 = 73.6%
L lo Jo.5
488 Chapter 4 Integration
f
Jo
TTt 2
80. I 1.75 sin ^;- A =
2 77
1.75 cos
f]'
--(1.75)(- 1 - 1) = - - 2.2282 liters
IT IT
Increase is
2 _ 5J_^ L9
77 TT TT
= 0.6048 liters.
J"' y3/2
Simpson's Rule (n = 4):
Jo
Grapliing utility: 0.166
£& =
■ 2(1/4)3/^ 2(1/2)3/^ 2(3/4)3/^ r
_ 3 - (1/4)2 + 3 _ (1/2)2 -^ 3 _ (3/4)2 "^ 2.
4(1/4)V2 2(1/2PA 4(3/4)V2 f
3 - (1/4)2 -^ 3 _ (1/2)2 3 _ (3/4)2 2.
== 0.172
= 0.166
Jo
84. Trapezoidal Rule (n = 4): J\ + %\v?xdx =» 3.820
Jo
Simpson's Rule (« = 4): 3.820
Grapliing utility: 3.820
Problem Solving for Chapter 4
2. (a) F(x) = sin r^ dt
r
X
0
1.0
1.5
1.9
2.0
2.1
2.5
3.0
4.0
5.0
F(x)
-0.8048
-0.4945
-0.0265
0.061 1
0
-0.0867
-0.3743
-0.0312
-0.0576
-0.2769
(b) G{x) =
x-2
i
sin f2 dt
X
1.9
1.95
1.99
2.01
2.05
2.1
G(x)
-0.6106
-0.6873
-0.7436
-0.7697
-0.8174
-0.8671
limG(x) = -0.75
(c)F'(2) = lim^^W^
i-»2 X — 2
= lim —
Jr-»2X
-2J2
sin t^ dt
= lim G{x)
i->2
Since F'{x) = sinx^, F'{2) = sin4 = lim G(x).
x^2
(Note: sin 4= -0.7568)
Problem Solving for Chapter 4 489
4. Let d be the distance traversed and a be the uniform acceleration.
We can assume that v(0) = 0 and s{0) = 0. Then
a{t) = a
v(r) = at
s{t) = -ai^.
s(t) = d when t = -./ — .
V a
The highest speed is v = a
/2d_
/ a
Jlad.
The lowest speed is v = 0.
The mean speed is -(j2ad + o) = .
f7d
V 2
The time necessary to traverse the distance d at the mean speed is
d fid
t =
Jadjl V a
which is the same as the time calculated above.
6. (a)
I I I I I I t I I I* <
0.2 0.4 0.6 0.8 1.0
(b) V is increasing (positive acceleration) on (0, 0.4) and (0.7, 1.0).
v(0.4) - v(0) _ 60 - 0
(c) Average acceleration =
0.4 - 0 0.4
(d) This integral is the total distance traveled in miles.
= 150mi/hr-
1
385
v(i) dt = -^[0 + 2(20) + 2(60) + 2(40) + 2(40) + 65] = ^ = 38.5 miles
Jo 1" 10
(e) One approximation is
,- -, _ v(0.9) - v(0.8) 50-40 ,^ ...2
(other answers possible)
490 Chapter 4 Integration
8. I mix -t)dt= \ xf{t) dt-\ tfit) dt = x\ fit) dt- \ tfit) dt
h Jo Jo Jo Jo
Thus, £ j f(t)(x -t)dt = xfix) + j fit) dt - xfix) = J fit) dt
Differentiating the other integral.
Thus, the two original integrals have equal derivatives.
\fit)ix -')dt=\ M /(v) dv\dt + C
Letting Jt = 0, we see that C = 0.
J"' 2 T 2
Vx dx = -x'''^ = -. The corresponding
0 3 Jo 3
Riemann Sum usmg right-hand endpoints is
5(n) =
'1
Thus, lim
= -^[^^ + 72 + • • • + v^]
yi + V2 + ■ • ■ + >/« 2
-,3/2
3'
12. (a) Area = ( (9 - x^) & = 2 I (9 - jc^) ate
= 2[27 - 9] = 36
(b) Base = 6, height = 9. Area = \bh = |(6)(9) = 36.
(c) Let the parabola be given by y = b^ — a^x^, a, b > 0.
rb/n
Area
rb,
= 2 it^- aV) dx
Jo
b^x - a^j^^
= 2
= 2
a 3 a
4^
3 a
2b
Base = — , height = b^
a
. .• . ,^ ■ . 2/'2fo\,,„ 4fc3
Archimedes Formula: Area = - — Hb^) = — —
3\ a / i a
4 -2-1 12 4 5
Problem Solving for Chapter 4 491
14. (a) (1 + /)' = 1 + 3( + 3r + P => (1 + i^ - i^ = 3P + 3i + 1
(b) 3r + 3; + 1 = (; + 1)^ - (3
J;(3/2 + 3,+ l)= 2[('+iP-'']
1=1 1 = 1
= (23 - V) + (33 - 23) + ■ • ■ + [((n + 1)3 - «3)]
= (« + 1)3 - 1
Hence, (n + 1)3 = J^(3r + 3/ + 1) + 1
i=l
(c) (n + 1)3 - 1 = 2(3,-^ + 3/ + 1) = 23,-2 + 3Mk±i) + „
1 = 1 1=1 -^
£3/2 = „3 + 3„2 + 3„ _
3n(« + 1)
2«3 + 6«- + 6« - 3n2 - 3/j - 2n
2«3 + 3?!- + n
n(/i + l)(2n + 1)
A., ^ w(^; + \)(2n + 1)
,^/ 6
16. (a) C
(b) C
/•20
= »'J.
12 sin
12 sin
vit - 8)
12
7T<r-8)
12
r 14.4 iT{t-
dt= cos —
L 77 12
-8)
-14.4
(-1 - 1) ^$9.17
A. \ 14.4 77<f-8) ^,,
j\dt = cos -r 0.6f
J L TT 12 Jio
14.4/ -V^
10.
H^
14.4/ 73
= $3.14
Savings = 9.17 - 3.14 = $6.03.
18. (a) Let A
Jo/U) +
fix)
f(b - x) ■
Let u = b - x, du = - dx.
, r f{b-u) . ^-
^ f /«^-«) ■
^ f /(fc-.x) ,
Jo/(^--^)+/W
Then,
2A
= f fix) , . f /(fc--v) ■
Jo/W + /(i - x) ^^ "^ Jo/(fc - .t) + fix) '^
Jo
\dx = b.
Thus, A = -.
(b)
Jo sin(l — j:) + sm.x 2
CHAPTER 5
Logarithmic, Exponential,
and Other Transcendental Functions
Section 5.1 The Natural Logarithmic Function: Differentiation .... 493
Section 5.2 The Natural Logarithmic Function: Integration 498
Section 53 hiverse Functions 503
Section 5.4 Exponential Functions: Differentiation and Integration . . 509
Section 5.5 Bases Other than e and Applications 516
Section 5.6 Differential Equations: Growth and Decay 522
Section 5.7 Differential Equations: Separation of Variables 527
Section 5.8 Inverse Trigonometric Functions: Differentiation 535
Section 5.9 Inverse Trigonometric Functions: Integration 539
Section 5.10 Hyperbolic Functions 543
Review Exercises 548
Problem Solving 554
CHAPTER 5
Logarithmic, Exponential, and Other Transcendental Functions
Section 5.1 The Natural Logarithmic Function: Differentiation
Solutions to Even-Numbered Exercises
2. (a)
(b) 3
The graphs are identical.
4. (a) In 8.3 = 2.1163
(b) -A = 2.1163
6. (a) In 0.6= -0.5108
(b) -df -0.5108
8. /(.x)= -Inx
Reflection in the .i-axis
Matches (d)
10. /W = -\n(-x)
Reflection in the v-axis and the .v-axis
Matches (c)
12. /(.v) = -2\nx
Domain: .ir > 0
14. fix) = \n\x\
Domain: x i^ Q
16. g(x) = 2 + In.T
Domain: .r > 0
18. (a) In 0.25 = In 5 = In 1 - 2 In 2 = - 1.3862
(b) In 24 = 3 In 2 + In 3 = 3.1779
(c) In ^12 = 5(2 In 2 + In 3) = 0.8283
(d) In^ = Inl - (3In2 + 21n3) = -4.2765
22. \nxyz = In.v + In v + In c
20. ln^^= In 2'/- = ?ln2
24. Inv a - 1 = ln(a - 1)"= = (;) ln(a - 1)
26. In 3e= = In 3 + 2 In e = 2 + In 3
28. In - = In 1 - In e = - 1
e
493
494 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
30. 3 Inx + 21ny - 41nz = \i\x^ + Iny^ - Inz!*
= ln
x'y^
32. 2[\nx - ln(jc + 1) - In^ - 1)] = 2 In
In
{x + l){x - 1) W - 1
X V
34. |[ln(;c2 + 1) - InU + 1) - \n(x - 1)] = |ln. /' t ^ ,.
In
V(i^
36.
3
/
38. lim ln(6 — ;c) = — oo
a:->6"
40. lim In— 7==
^-►5* Vx - 4
ln5 = 1.6094
42.
>> = Inx
, 3
3/2
= -Inx
At (1, 0),
y'
_ 3
~ 2-
44. >- = lnx'/2 = iln^:
•' 2x
At(l,0),>'' = i
46. /iW = ln(2x2 + 1)
1 \x
2x2 + r ^ 2jc2 + 1
48. y = ;c In X
^ = 4-) + Inx = 1 + Inx
dx \xJ
50. >- = lnVx2 - 4 = - InU^ - 4)
rfy _ 1/ 2r
a[r 2\jc2 - 4/ x^ - 4
52. f(x) = In
/'(x) =
2x
,x + 3
1 1
In 2x - ln(x + 3)
3
X x + 3 x{x + 3)
54. h(t)
h'(t)
\nt
ti\/t) -\nt \ -Int
56. y = In(lnx)
dy _ lA _ 1
lie \nx a: In x
58. y = In 3/^ = h^i^ " D " ln(^ + D]
X + 1 3'-
, ^ ir_! !_
' 3U - 1 x+ \
1 2 2
3 ^2 - 1 " 3(a:2 - 1)
60. /W = \n{x + V4 + x^)
1
/'W =
x+ V4 + x2 V x/4T
1 +
1
-.--c ^^5
-i.^5
> ^ -
^' '"
c-Jc-f
-.;c<ef<
Section 5. 1 The Natural Logarithmic Function: Differentiation 495
62. > = ^ 4ln( ) = —^^ -\ril + JJT^) + -\nx
1
H
^ ^ -lx'^{x/Jx- + 4) + 4;cV;c^ + 4 _ 1/ 1
dx 4x* 4^2 + Vxn^J\V^^^T4] ' 4x
Note that:
1
1
2 - y;c- + 4 2 - Vx- + 4
2 + Va;2 + 4 2 + 7.x:2 + 4 2 - Va:= + 4
-;c2
Hence, -i- =
-1
y^M^ 1 (2 - Jx^ + 4)
dx " IxJx^ + 4 x^ 4 -.t2
-1 + (l/2)(2 - V?T4) ^ y/jc^ + 4 ^ 1
Ixjx^ + 4 .t^ 4j:
Vx^ + 4/ ' 4x
- V.t- + 4 Vj:' + 4 J_ ^ Jx^ + 4
4;c7;c- + 4 -T^ 4j: .x^
64. >! = ln|csc;<:|
— CSCJ: • cot AT
y
cscj:
= -cotx
66. 3; = Inlsecx + tan.i;|
dy _ sec .r tan .r + sec'.r
dx sec .r + tan -T
sec .i:(sec x + tan x)
sec a: + tan x
secx
68. y = InVl + sin^.x: = -ln(l + sin^x)
dy /'1\2 sin.i:cos;c sin j: cos x
dx \ll 1 + sin-x 1 + sin-.r
70. ^(.r)
/•Ihj:
+ 3)rff
g [x] = L(ln.r)- + 3J— (ln.r) =
dx X
(Second Fundamental Theorem of Calculus)
72. (a) y = 4 - x2 - Inl -X + 1 ), (0. 4)
^=-2;, \ i
dx (l/2).r + 1 V2
= -2jc-
1
x + 2
When.x: = 0,^= -\.
dx 2
Tangent line: >• - 4 = - -[x - 0)
y = --.r + 4
74.
ln(.xT) + 5.r = 30
In.T
+ In y + 5.r = 30
1_
x
+ 1^ + 5 = 0
ydx
\dy _ 1
y dx X
- 5
dy_ y
dx X '
- 5y = -
-F^
(b)
496 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
76.
y = j:(ln x) — Ax
y' = x\-\ + \nx — A = — 3 + lnj:
(x + y) — xy' = X + x\nx — Ax — x(—?) + \r\x) = Q
1%. y = x - \nx
Domain: x > Q
V ' = 1 = 0 when x = I.
X
/'=-> 0
X-
Relative minimum: (1, 1)
80. v =
Injc
Domain: x > Q
, _ x{\lx) -\nx _ 1 — Inj:
= 0 when x = e.
u.-')w.¥^)
„ ^ ^{-\lx) - (1 - ln;c)(2x) ^ 2(lnx) - 3
Relative maximum: (e, e~ ')
Point of inflection: {e'^l^,\e''^l'^
0 when x = e^l'^.
82. y = x^ln -. Domain j: > 0
y = j:'I-I + 2j:ln- = jcI 1 + 2in-l = 0 when
-1 = 21n =
In-
x = 4e-i/2
/'=l+21n^ + 2x(y = 3 + 21n|
y"= 0 when;c = 4€-3/2 ' '
Relative minimum: (4e" '^^, - 8e" ')
Point of inflection: (Ae'^^^, -lAe'^)
4(4f-3'2.-24e-^
84. /(x)=xln^ /(I) = 0
f'(x) = 1 +ln;c, /(1)= 1
fix) = ^. /"(I) = 1
P,W =/(l) +/'(1)U - 1) = ;c - 1, P,(l) = 0
P^ix) = /(I) + f'{\)(x - 1) + |/"(1)U - 1)^
= ix- 1)+|U- 1)^ ^2(1) = 0
/•,'« = 1, P,'(l) = l
Fz'W = 1 + U - 1) = ;c, PjXl) = 1
The values off, Pj, Pj, and their first derivatives agree at
X = 1. The values of the second derivatives of/ and Pj
agree at;c = 1.
\P2
/
V
Section 5.1 The Natural Logarithmic function: Differentiation 497
86. Find x such that In x = 3 - x.
fix) = X + (In a:) -3 = 0
n
1
2
3
Xn
2
2.2046
2.2079
fUn)
-0.3069
-0.0049
0.0000
Approximate root: x = 2.208
90. y
\ X- +
- 1^
1
\r
lny = ^lnU2- I) - \n{x^ + 1)]
}_dy ^ I
y dx 2
2x 2x 1
.r= - 1 .v2 + ij
^ = A^ - 1 r 2x 1
ate " V.)c2+ iLx^- ij
U" - l)'/-2x
(.X^ + l)'/2(;c2 - 1)(.^2 + 1)
(x^ + iy'~{x- - 1)
1/2
88. y = yU - \){x - 2)(;c - 3)
\ny = ^[\n(x - 1) + ln(;c - 2) + ln(jc - 3j]
i(^uir_L_ + _^ + _L_i
y\dxl llx - \ X - 2 x-3]
^ If 3j:- - 12x + 11 1
2|_U- l)U-2)(x-3)J
dy ^ 3x' - lit + 11
dx 2v
3x- - 12x + 11
2VU - 1)U - 2)U - 3)
92. y
{x +\){x + 2)
ix - l){x - 2)
Iny = InU + D + InU + 2) - ln(.t - 1) - Hx - 2)
1 1
+
1
1
Udy
y\dxj X + \ X + 2 x — \ X — 2
dy
dx
XT — I X — 4
-6x- + 12 ]
■^ - DCr^ - 4) J
(x-
-6(x^ - 2)
(x + l)(x + 2)
ix - \)ix - 2) ix+ l)(x - l){x + 2)U - 2)
6(x^ - 2)
U - D^Ct - 2)2
94. The base of the natural logarithmic function is e.
96. gix) = ln/(x),/(x) > 0
.fix)
g'(x)
fix)
(a) Yes. If the graph of g is increasing, then ^'(.v) > 0.
Since/(x) > 0, you know that/'U) = g'ix)fix) and
thus, fix) > 0. Therefore, the graph of/ is increasing
(b) No. Let/(.r) = .r- + 1 (positive and concave up), gix)
Intxr + 1) is not concave up.
98. t
5.315
6.7968 + In.v'
(a) 50
1000 < x
(b) f(1167.41) = 20 years
T= (1167.41)(20)(12) = $280,178.40
(c) f(1068.45) ^ 30 years
T = (1068.45)(30)(12) = $384,642.00
(d) ^= -5.315(-6.7968 +
dx
5.315
.,,.(1)
.v(- 6.7968 + ln.v)=
Whenx = 1167.41. dt/dx « -0.0645. When.r = 1068.45,
dt/dx=- -0.1585.
(e) There are two obvious benefits to paying a higher monthly
payment;
1. The term is lower
2. The total amount paid is lower.
498 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
100. (a) 350
_, ,„ , 34.96 ,3.955
(b) T (p) = + — 7=^
P Jp
r'(10) = 4.75 deg/lb/in^
r'(70) = 0.97 deg/lb/in^
(C) 30
lim r'(p) = 0
102. >> = lOlnf^^"^ ^^°° — ^j - VlOO - x^ = 10[ln(l0 + VlOO - ;c2) - Inx] - VlOO - x^
(a) ?2 (c) lim ^ = 0
j:->10- dX
dy
Vioo - xKio + Vioo - x^)
-a
yioo"
-10
yioo"
10 + Vioo - x^.
-10
-10 +
Vioo - jc'
Vioo - r
10+ Vioo - x'^
Vioo - x^
+ 1
10 + Vioo - x^
10
X Vioo - x^
10
x
w
x
10 + Vioo - x2 Jc
;c(lO - VTOO^^ - 10 ^ Vioo - x^
When x = 5, dy/dx = - Vs. When x = 9, (fy/dc = - Vl9/9.
104. y = in X
y ' = - > 0 for jc > 0.
Since In x is increasing on its entire domain (0, oo), it is a
strictly monotonia function and therefore, is one-to-one.
106. False
TT is a constant.
d,
dx
[In tt] = 0
Section 5.2 The Natural Logarithmic Function: Integration
\ — dx= \o\-dx= lOlnj
x\ + C
4. u = X — 5, du = dx
1
/.
:& = Inx - 51 + C
\z^2'' = \h
2'' = -2kT2^'^'^
'x- 5
S. u = 3 - x^, du = — 3x'^ dx
= -In |3x + 2| + C
Js^'^^-l/l^^-^x^)^
--ln|3 -x^\ + C
Section 5.2 The Natural Logarithmic Function: Integration 499
10. M = 9 - x\du = -2xdx
J V 9 - x^ 2J
r xU + 2) if 3;c- + 6^:
J ^ + 3x2 - 4 3J ^ + 3^2 _
dx (u=x^ + 3x' - 4)
^ln|x3 + 3x2-4| + C
,^ r2;r + 7A-3^ f
7a- - 3 ^ I /^ , , 19 , _,
-dx = 2x + 1 1 + dx
x-2)
= x^ + lU + 191nU - 2| + C
18.
fx^ - 3;c2 + 4;c - 9 , f
J — ^^n — '^ = J
-3 + .1: +
a:- + 3
a[t
a;2 1
= -3;t + y + -lnU= + 3) + C
-1^^^- = /
^-^K-— "-^i-
3 2
+ 19;c- n51nU + 5| + C
JxlnCr^) 3 Jinx x
J ln|ln|x|| +C
22. M = 1 + x'/3 du
3x2/3
a^
jx2/3(l +x'/3)'^ Y
/3)'^ = 3jY^(^j^
31n|l +x'''3| + c
= Inlx - ll +
1
2(x - 1)
- + c
26. M = 1 + V3x, rfM = -A=dx ^ (& = -(«- 1) rf«
2v'3x 3
J 1 + V3I J«3
l)d«
rfM
^M - ln|M|] + C
= 1^1 + V3^ - ln(l + v/3l)] + C
= |v/3x-|ln(l + V3^) + C,
500 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
/:
'x- 1
28. u = x'/3 -\,du= T-^ dx => dx = Z{M + Xfdu
dx = 2{u+\)^du
= 3 f ^^^(«' + 2« + 1) du
= 3 ( M^ + 3« + 3 + -j rf«
= 3rj + ^ + 3M + lnlM|l + C
3^
3 2
1/3 - 1)3 , SU'/J - 1)2
+ ■
+ 3(;c'/3 - 1) + ln|x'/3 - l|
+ C
3x2/3
= 3 ln|;c'/3 - 1 1 + =^ + 3;c'/3 + x + Q
9. \tan5ede = -
30. |tan5erfe = Tl^^rffl
cos 56
= --ln|cos5e| + C
32. I sec - (fe = 2 I sec -( - j <ic
= 21n
x , X
sec - + tan -
2 2
+ C
34. M = cot t, du = — csc^ f A
rcsc-^ t
f-
cot f
-rf; = -In cot f + C
/'
36. (sec f + tan r) A = Inlsec r + tan fl - Inlcos t\ + C
= ln
sec r + tan f
cos t
+ C
= ln|sec t(sec f + tan r)| + C
38. y
h
2x
dx
= ln|;c2 _ 9| + c
(0,4): 4 = ln|0- 9| + C
)' = ln|;c2_ 9| +4- in9
C = 4 - ln9
)
■^^ (0, 4)
^
7
V
40. , = f-J££il-
J tan f + 1
dt
= ln|tanf + 1| + C
(77,4): 4 = ln|0+ 1| + C =» C = 4
r = Inltant + ll + 4
w
^
<T.'4)'
42.^ = ^,(1,-2)
(a) y
X 2
(In 1)2
+ C ^ C= -2
Hence, y = - 2.
Section 5.2 The Natural Logarithmic Function: Integration 501
^•£7T3'^=K^2|]-
In 3 - In 1 = In 3
46. u = \n X, du = — dx
X
J^ x\nx j^ \lnxjx
In InU
= ln2
48. f^— |-A= I 1^+ |— TT^
Jo j: + 1 Jo Jo ^ + 1
[
= U - 21nl;c + ll
1 - 2 In 2
J -0.2 ro.2
(esc 20 - cot ley de = (csc^ 20-2 esc 20 eot 20 + cot^ 20) dd
0.1 Jo.i
ro.2
•20- 2ese20cot20 - I) dO
J -0.2
(2 esc2
0.1
= -cot 20 + esc 20 - 0 = 0.1
0024
52. ln|sinx| + C = In
+ C = -ln|eseA:| + C
54. -Inlesej: + cotJ:| + C = —In
-In
(escj: + cot x)(ese x - eot x)
(esc ;c - cot x)
+ C = -In
csc-.t - eot"^.x
esc j: — eot X
+ C
1
cscx - eotx
+ C = In [esc x - cotxl + C
56.
\- ^dx = -(l + V^f + 6(l + V^) - 41n(l + v^) + Ci
J 1 + V x
= 4v^ - X - 4 ln(l + v^) + C where C = C, + 5.
58. — dx = -rflnlsec Zx + ian2x\ — sin 2x] + C
ps
J-7r/4
^. , sin"^x - eos'j: ,
60. dx
[ini
secx + tanx - 2 sinx
ir/4
-ir/4
Inl'^^'l - 2v^ - - 1.066
Note: In Exercises 62 and 64, you can use the Second Fundamental Theorem of Calculus or integrate the function.
62. F{x)
F'{x) = tanx
= I tantdt
Jo
64. F{.x) = j^^
dt
Zx 2
F\x) = 4 = -
.r- .V
502 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
66.
A = 3
Matches (a)
68. A
r^-i
X + 4 In ;
>.]'
4 + 4 In 4 - 1
3 + 4 In 4 » 8.5452
••I
70. {2x - tan(0.3x)) dx
x'- + ■
10
ln|cos(0.3x)|
10,
16 + Ylncos(1.2)
]-hf
In cos(0.3)
« 11.7686
72. Substitution: (m = x^ + 4) and Power Rule
74. Substitution: (« = tan ;c) and Log Rule
76. Answers will vary.
78. Average value =
1 r4(x-
4 - 2J2 ^2
4., J
+ 1)
dx
+ -z\dx
X x-^'
re
82. f
I
= 2
1
In 4-;^- In 2 + 1
ln2 + ^l = ln4 + ^« 1.
8863
10
In 2 ,„ r - 100
dT
80. Average value
10
In 2
ln(r - 100)
= i^[.n200-lnl50] = i^[ln(|;
2-oJo
TTX ,
sec-— dx
6
l/'6
In
TTJC . TTX
sec -^ + tan — -
6 6
L2\Tr,
= -[ln(2 + 73) - ln(l + 0)]
= -ln(2+ 73)
=» 4.1504 units of time
84.
dS^k
dt ~ t
S{t) = \ -dt = k\n\t\ + C = klnt + C since r > 1.
5(2) = A: In 2 + C = 200
S(4) = k\n4 + C = 300
Solving this system yields k = 100/ln 2 and C = 100. Thus,
5W - -!5^ + 100 - 1001
[^-1
Section 5.3 Inverse Functions 503
86. k= I: /iW = X- \
Vx- 1
k = 0.5: /0.5W =
A: = 0.1:/o.,W
lim L (x) = \nx
;t-.0*
0.5
0.1
= 2(v^ - 1)
= io('^- 1)
88. False
— [Inx] = -
ax or
90. False; the integrand has a nonremovable discontinuity at
x = 0.
Section 5.3 Inverse Functions
2. (a) f(x) = 3 - 4a:
/(.W)=/f^)-3-4i^Ux
'g(/W) = g(3 - 4x) = ^ ^l ^-"^ = X
(b)
/\..
4. (a) /W = 1 - x' •
f(g{x)) = /( yp^) = 1 - ( yr^^)
= 1 - (1 - x) = JC
g(/W) = gil - x')
= Vi - (1 - ;c^) = l^ = .t
6. (a) /W = \6 - x\ x > 0
g(x) = 716^^
/(gW) =/(yT6^^) = 16 - (VT6^^)2
= 16 - (16 - x) = X
g{f(x)) = g(16 - ;c2) = Vl6 - (16 - x^)
(b)
8 12 16 20
= vj: = JC
8. (a) f(x) = — — , .X > 0
1 + X
g(x) = ^-^^, 0 < .X < 1
f{g{x))=f
gifix)) = gl
x
1 -.V
1 1
X X
1
1 + X
1 + -Y _ X 1 + -V
1 ~ 1 + .X- ' 1
1 +.r
(b)
504 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
10. Matches (b)
14. /U) = 5.r- 3
One-to-one; has an inverse
12. Matches (d)
16. F(x) =
x^ + A
Not one-to-one; does not have an
inverse
18. git)
1
Jt^ + l
Not one-to-one; does not have an
inverse
20. fix) = 5x^x - 1
One-to-one; has an inverse
22. hix) =\x + 4\ - \x-4\
Not one-to-one; does not have an inverse
24. fix) = cos-
3x
, . 3 . 3j: Itt 4tt
fix) = —— sin—- = 0 when x = 0, -r-, ^r-, ...
•' 22 33
/is not strictly monotonic on (— oo, oo). Therefore, /does not have an inverse.
26. fix) = j^ - 6^2 -t- 12x
/'W = 3;c2 - 12t -K 12 = 3(;c - 2)^ > 0 for all x.
/is increasing on (-oo, oo). Therefore, /is strictly
monotonic and has an inverse.
28. fix) = ln(x- 3),;c > 3
1
fix)
x-3
> OfoTx > 3.
/is increasing on (3, oo). Therefore, /is strictly monotonic
and has an inverse.
30. fix) = 3x = y
"I .
X
/-w = f
32. fix) =x^ - 1
x = ^y + 1
y = ^x + 1
r'W = ^^TTT
34. fix) =x^ = y, 0 < X
X = Vy
y = Vx
f-'ix) = ^
Section 5.3 Inverse Functions 505
36. fix) = Jx^- A = y, x>2
X = ^y^ + 4
y = Jx^ + 4
/-'W = V.r^ + 4, jc > 0
38.
fix) = 3 V2x - 1
y5 + 243
^ 486
.t^ + 243
^ 486
j^ + ''43
486
h^
C^
1
The graphs of/and/~'
are reflections of each
other across the line v =
40. fix) = a:V5 = y
V = ^5/3
/-'W = Ar5/3
/^
^
The graphs of f and/" ' are
reflections of each other across
the line v = x.
X + 2
42. fix) = ^-^ = V
.r
/-'W =
y- 1
2
j: - 1
2
;c - 1
■ — : —
^
1
\
44.
X
0
2
4
/-'W
6
2
0
The graphs of /and/"' are
reflections of each other across
- .. • the line y = x.
46. fix) =ki2- X- X?) is one-to-one for all k ^ 0. Since/-'(3) = -2,/(-2) = 3 = ^2 - (-2) - (-2)3) = 12A.- => k = \.
48. fix) = \x + 2| on [-2,00)
/'W=^^l) = l >0on(-2,oo)
X + 2
/is increasing on [-2, 00). Therefore, /is strictly
monotonic and has an inverse.
52. fix) = secjcon
0.-
50. fix) = cot -t on (0, tt)
fix) = -csc-.t < Oon (0, tt)
/is decreasing on {0, tt). Therefore, /is strictly monotonic
and has an inverse.
fix) = sec X tan x > 0 on I 0, — I
/is increasing on [0, it/2). Therefore, /is strictly monotonic and has an inverse.
506 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
54. /(.r) = 2-^ = yon(0, 10)
2^2 _ 3 = y.Zy
xK2 - y) = 3
f-'{x)
56. (a), (b)
= -V2
3
~ y
= *V2-.
2
K'
/^.
12
The graphs of/ and/"' are
reflections of each other
across the line y = x.
58. (a), (b)
(c) h is not one-to-one and does not have an inverse.
The inverse relation is not an inverse function.
(c) Yes, /is one-to-one and has an inverse. The inverse
relation is an inverse function.
60. /(x) = -3
Not one-to-one; does not have an inverse
64. f(x) = 16 — j:^ is one-to-one for j: > 0.
16-;c^ = y
16 -y = x^
4/16^ = ^:
Vl6 - x = y
f-\x) = yi6 - X, X < \6
62.
f{x) =
ax + b
/is one-to-one; has an inverse
ax
+ b
= y
X
y-b
a
y
X - b
a
f-
-'W
-'-\a^Q
a
66.
/w =
1^-
3| is one-to-one for X > 3.
x
- 3 =
= y
X =
--y + 3
y
-x + 3
f-
Kx)-
= X -1- 3, X > 0
68. No, there could be two times t^ i= t-^^ for which
hit,) = h{t^.
70. Yes, the area function is increasing and hence one-to-one.
The inverse function gives the radius r corresponding to
the area A.
72. fix) = ^ix? + 2x3);/(-3) = ^("243 - 54) = - 11 = a.
1
fix) = ^5x^ + 6x2)
(/-')'(- n)
1
1
27
^ ^ ^\_
7'(/-'(-ll)) /'(-3) 5(-3)^ + 6(-3)2"l7
Section 5.3 Inverse Functions 507
74. f{x) = cos 2x,/(0) = 1 = a
f'{x) = -2sin2x
^•^"'^'^'^ =7lrW) =7W = ^i^ = ^ which is undefined.
76. /W = v^r^^,/(8) = 2 = a
1
Z'U)
(/-')'(2)
2jx- 4
■/'{/- '(2)) /'(8) 1/(278^^) 1/4
78. (a) Domain/ = Domain/ ' = (-00,00)
(b) Range/ = Range/"' = (-00,00)
(c)
80. (a) Domain / = [0, 00), Domain /-' = (0, 4]
(b) Range / = (0, 4], Range /"' = [0, 00)
(c)
(d) /(.r) = 3 - 4x, (1, - 1)
fix) = -4
/'(I) = -4
f-\x)=^. (-1,1)
(r')w = -^
(/-')'(-!) =4
(d) /w=i',.
f'(,A ~<iX „ , ^
/(->) (^ + „r /(I) 2
rw=7^
— 9
1
2
1 2 3
82. x = 2 ln{y^ - 3)
1 = 2-
1
.rfy
^^2^"^
dtc 4y ■A'^"''^^'^^ 16 16-
In Exercises 84 and 86, use the following.
fix) = ix-3aQdg{x) = x-^
J^\x) = S{x + 3) andg-Hx) = ^
84. (g-' o/-')(-3) = r'(/-'(-3) = g-'(O) = 0
86. (g-' og-')(-4) = g-'(g-H-4)) = g-il-4)
= V^^= -^4
508 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
In Exercises 88 and 90, use the following.
fix) = X + 4 and g(x) = 2x - 5
J^\x) =x-4 andg-\x) = ^^
90. (g "/)(x) = gifix))
= gix + 4)
= 2(x + 4) - 5
= 2a: + 3
88. if-'
'g-')(x)=f-'(g-'ix))
■ -rm
=^-
x-3
Hence, {g »/)"'«
x-3
(Note:(go/)-'=/-'og-')
92. The graphs of/ and / ' are mirror images with respect
to the line y = x.
94. Theorem 5.9: Let/be differentiable on an interval /.
If/ has an inverse g, then g is differentiable at any x for
which f'(g{x)) + 0. Moreover,
g\x)
1
/'(gW)
. f\g{x)) + 0
96. /is not one-to-one because different j:- values yield the
same y- value.
Example: /(3) = /
Not continuous at ±2.
98. If/ has an inverse, then /and/ ' are both one-to-one.
Let (/-')~'W = y then x = r\y) and/W = y.
Thus, (/-•)-' =/
100. If/has an inverse and/(xi) = f(x-^, then/"'(/Ui)) =/~'(/U2)) =^ ^i ~ ^2- Therefore,/is one-to-one. If/W is one-to-
one, then for every value b in the range, there corresponds exactly one value a in the domain. Define g{x) such that the
domain of g equals the range of/ and g(Z?) = a. By the reflexive property of inverses, g = /"'.
102. True; if/has ay-intercept.
104. False
Let/U) = xoxg{x) = \/x.
106. From Theorem 5.9, we have:
,. ,_fUx)m-f"{g{x))g\x)
ng{x)) ■ [i/r(gU)))]
\f'ig{x)W
figix))
\fUxm
If/is increasing and concave down, then/' > 0 and/" < 0 which implies that g is increasing and concave up.
Section 5.4 Exponential Functions: Differentiation and Integration 509
Section 5.4 Exponential Functions: Differentiation and Integration
2. e-2 = 0.1353.
lnO.1353. .. = -2
8. 4^^ = 83
83
e'
X = ln(f
- 3.033
12.
200e-
-4c —
15
e~
-4i =
15 3
200 40
^
4x =
■»(i)
x =
i'"(f)'
0.648
16.
ln4x
= 1
4a:
= e'
= e
a;
e
" 4
- 0.680
In 0.5 = -0.6931...
6.
gta2x= 12
0.6931. . . = i
2;c = 12
X = 6
10. -6 + 3^^ = 8
ie' = 14
-^
-
. = ln(^
) =
= 1.540
14. In;c2=10
^2 = glO
= ±e' ^±148.4132
18. Hx - 2)2 = 12
(x - IT' = e>2
X - 2 = e«
x = 2 + e^ == 405.429
20. V = W
22. V = e"^/2
24. (a)
(b)
"^
V.
Horizontal asymptotes: y = 0 and y = S
Horizontal asymptote: v = 4
510 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
26. y = Ce"""
Horizontal asymptote; y = 0
Reflection in the y-axis
Matches (d)
28. y =
1 +e-
lim
j^oo 1 + «-"
lim -z
x-*-ca I + e
= c
0
Horizontal asymptotes: y = C and y = 0
Matches (b)
30. fix) = e^/3
g(x} = In x^ = 3 In jc
32./W = e--'
g(x) = 1 + Inj:
34. In the same way.
lim
(-3-=
e^ for r > 0.
,^,,111 1 1 1
36. l + l+x + T + :rT + ttt + ^rr +
2.71825396
2 6 24 120 720 5040
e == 2.718281828
.>l + l+^ + ^ + :^ + 4. + a.+ 1
2 6 24 120 720 5040
38. (a) y = e^
y' = 2e^
At(0, l),y'= 2.
(h) y = e-^
y'= -2e-^
At (0,1),)''= -2.
40. /W = e'-^
fix) = -e'--
42. >> = e
= ^-^
^=-2x.-^
44. v = x^e
dy
dx
—x^e ^ + 2xe ^
= xe-'{2 - x)
46. g(/J = e-^'^
g'it) = e-V/^Cer-^)
S^S/r'
fe
48. >> = Ini
1 + e'
1 - e^,
ln(l + e') - ln(l - e')
+ ■
dr 1 + e' ■ 1 - e'
2e^
1 - e
2x
50. > = Inl
e^ + e"
In(e^ + g-^ - In 2
dy _ e^ — e ■
dx e^ + e~-
e^ - 1
e^ + 1
52. y =
e' — e
^~ 2
54. y = xe^ — e' = e'ix — 1)
dx
= e' + e'ix - \) = xe^
Section 5.4 Exponential Functions: Differentiation and Integration 511
56. f(x) = e^\nx
fix) = -
X
58. y = \ne' = X
dx
60. e^ + X? - f = 10
\x^ + y\e^ + 2x - ly^ = Q
\ dx dx
di
dx
(xe^ - 2y) = -ye^' - 2x
dy
dx
xe^ - ly
62. g{x) = -Jx + e'Xnx
1 e'
— P + — + e* In j:
,,, . 1 xe' — e' e'
1 , e^jlx-l) , ,,
- H :: h e' In jc
4xv^
;c2
64. y = ^^(3 cos 2x - 4 sin Ix)
y' = e^i-6smlx - 8 cos 2x) + f^(3 cos 2x - 4 sin 2x)
= e^{—\Qsix\lx - 5 cos Ix) = - 5e'^(2 sin Ix + cos 2x)
y" = - 5e'(4 cos 2x - 2 sin 2j:) — Se'il sin 2j: + cos Ix) = - 5e^(5 cos 2j:) = - 25^^ cos 2x
y"- ly' = -25e'cos2jc - 2(-5e')(2 sin 2j: + cos 2;c) = -5e*(3cos2x - 4 sin 2j:) = -5y
Therefore, y" - ly' = - Sy => y " - 2y ' + 5.v = 0.
/■
66. f{x) =
e' — e ^
/'W = ^-^ > 0
/W
oX — a X
= 0 when x = 0.
Point of inflection: (0,0)
(0.0)
/
A
68. g(x) = ^e-(--3)V2
^'W = -^(;c-3>-(-3)V2
^2^
5"W
1
/27r
Relative maximum
{x - l){x - 4)e-(^-3)V2
^ ('•*)
(3, 0.399)
Points of inflection: I 2, ^=«- '/^ |, ( 4, — L=^- '/:
- (2, 0.242), (4, 0.242)
70. fix) = xe-''
f'(x) = -A:e~-' + e~' = e"^(l -;«:) = 0 when.v = 1.
f"(x) = -e--^ + (-e-')(l - x) = e-'ix - 1) = Owhenx = 2.
Relative maximum: (l.e'')
Point of mflection: (2, 2e-2)
(I..--')
/
512 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
72. fix) = -2 + e3^(4 - 2x)
fix) = eH-2) + 3e3^(4 - 2x) = e'^ClO - 6x) = Owhen;c = f.
fix) = e^1-6) + Se^HlO - 6x) = £^^(24 - 18;c) = Owhen;c = f.
Relative maximum: (5, 96.942)
Point of inflection: (|, 70.798)
74. (a) fie) = fie + x)
lOce"'^ = 10(c + ;c)e-<''+^)
C_ _ C + X
ce"^" = ic + x)e''
ce' = c + X
ce^ — c = X
X
(c) Aix) = -J^^/»-^*
e^ — 1
(b) AW = jcf (c) = X
10'
10x2
e^- 1
e^ - 1
-(V(e'-i)l
e^/d-c')
(d) c =
e'- 1
lim c = 1
lim c = 0
The maximum area is 4.591 for a: = 2.118 and
fix) = 2.547.
76. Let (j;o> y^) be the desired point any = e ^.
-e " (Slope of tangent line)
1
-, = e' (Slope of normal line)
y — e~'^ = e^'ix — x^
We want (0, 0) to satisfy the equation;
f( 0.4263, ^-<'«")
1 = x^e^
XqC^ -1=0
Solving by Newton's Method or using a computer, the solution is Xq = 0.4263.
(0.4263, e-o"2«3)
Section 5.4 Exponential Functions: Differentiation and Integration 513
78. V= 15,0{)Oe-o«286, 0 < f < 10
(a) 20.000
dV
(b) ^ = -9429e-o«286'
at
Whenr =1,^= -5028.84.
dt
Whenr = 5,^- -406.89.
dt
(C) 20.000
80. 1.56e ''■^cos4.9r < 0.25 (3 inches equals one-fourth foot.) Using a graphing
utility or Newton's Method, we have f > 7.79 seconds.
'O'V
82. (a) V, = -1686.79f + 23,181.79
V^ = 109.52f2 - 3220.12t + 28,110.36
(b) The slope represents the rate of decrease in value of
the car.
(c) V3 = 31,450.77(0.8592)' = 31,450.77e^'"5i8t
dV,
(e) -r^ = -4774.2e-
dt
dV^
For f = 5, — ^ = -2235 dollars/year.
dV,
For f = 9, — ^ = - 1218 dollars/year.
(d) Horizontal asymptote: lim V3(r) = 0
As f — > 00, the value of the car approaches 0.
/"l
-^
\^'
/
\
84. fix) = e--^/2,/(0) = 1
fix) = -xe-^'\f'{Q) = 0
fix) = .t^e-^/2 - e-^/2 = e-^/2(;c2 - l),/t0) = - 1
P.W = 1 + 0(.t - 0) = 1, Pi(0) =1
P, '(.v) = 0, P, '(0) = 0
P^ix) = 1 + 0(.r - 0) - i(.r - 0)= = 1 - y, P^lO) = 1
P^'ix)^ -x,P2'{0) = 0
P.'\x) = -UP, 'to) = -1
The values of/, P,, P2 and their first derivatives agree at ;c = 0. The values of the second derivati\es of
/and P2 agree at jc = 0.
514 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
86. n* term is x"/n\ in polynomial:
r= x^ ;c^
X^ JC'
Conjecture: e^ = 1 + x + — + — +
»■/:
90. J e^-'dx
-e-i + 1 = 1-
>
88. Let M = -x^, du = -Ax^dx.
3^
e "\-Ax^)dx = e "' + C / ' vj
v.. ^~
^
c.
7
92.
[:t2^/2^ = ll
jt ->,' ' J ^
^^^(¥) '^ " \^"' "^ *^
J^d.= -ije
94. r^dx=-\\e^'4-^)dx=-\e^'^ + C
96. Let M = 1 + e^JT, rfw = le^dx.
+ e^A:) + c
98. Let u = ~z~. du = —x dx.
rV2 r.
xe-'^^-dx = -
Jo Jo
xe -''-dx = - f %--=/2(-x) a[x = [-^-^2!^-^=
1 -e-' =
e - 1
100. Let M = e^ + e^^ ^m = (e' - e-^)a[r.
/
e' + e'"
dx = ln(e^ + g-^) + C
102. Let M = e^ + e-", du = {e^ - e'") dx
- le
-2
e^ + e"
+ C
104. ^ ^ alt = (e^ + 2 + e-'^)dx
= e* + 2a: - e"^ + C
/'
106. le'^^^seclxtanlxdx = -e'^^ + C
(u = sec 2x,du = 2 sec 2j: tan 2x)
ln(e2»-i)a[x= I (
108. \\nie^-'^)dx = {2x - \)dx
x^ - X + C
112. f'{x) = I (sin;c + e^ a^ = -cosx + ^e^ + Q
/'(O) = -1+1+ C, =1^ C, = 1
e-^Pa[r
(e^ - 2 + e^2')a[«
= |e^ - 2;c - ^e~^ + C
f'(x) = -cosx + -e^+ 1
/W
= /
- cos x + -e^ + I ]dx
= - sin ;c + -e^ + x + C,
4 '^
/fO) = - + Q = -^Q = 0
/(x) == X - sinx + -e^
4
Section 5.4 Exponential Functions: Differentiation and Integration 515
114. (a)
(b)
di
dx
-0.21^
y = \xe-°^dx = —- \e-°^ {-OAx)dx
-0.2^
0.4
+ C = -2.5e-
+ C
0, -|]: -|= -2.5e'> + C= -2.5 + C =^ C=l
y = -2.5e-°-'' + 1
116.
f
Jn
e ' dx =
118.
f
Jo
(e-^ + 2)dj:
1
1 _.
+ 4 + 1 = 4.491
Jo
120. (a) Jxe" dx,n= 12
Midpoint Rule: 92.1898
Trapezoidal Rule: 93.8371
Simpson's Rule: 92.7385
Graphing Utility: 92.7437
/:
(b) 2x6-" dx,n = 12
Jo
Midpoint Rule: 1.1906
Trapezoidal Rule: 1.1827
Simpson's Rule: 1.1880
Graphing Utility: 1.18799
122.
\ 0.3-°^' dt = \
Jo
+ 1 =
-0.3 J = ±
2
-0.3a: = ln-= -In 2
ln2 ,,, .
X = —I- ~ 2.31 mmutes
124.
t
0
1
2
3
4
R
425
240
118
71
36
In/?
6.052
5.481
4.771
4.263
3.584
(a) InR = -0.6155f + 6.0609
K = g-0.6155; + 6.0«)9 = 428.78^""°*"^'
(b)
(c)
R(t)dt =
Jo Jo
428.78€
-0.61551 J, -.
dt = 637.2 liters
516 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
126. The graphs of f{x) = In j: and g{x) = e^ are mirror
images across the line y = x.
128. (a) Log Rule: (m = e^ + 1)
(b) Substitution: (m = x^)
130. \n-r = ]xie' -\ne^ = a- b
Ine""* = a - b
Therefore, In -r = In e" * and since y = In x is one-to-one, we have -r = e"
Section 5.5 Bases Other than e and Applications
Atro=16,y = y =-
-=(r
Atto = 2,y=(|)'^'-0.7579
6. log2,9 = log2,272/3=|
8. log„ - = log„ 1 - log„ a = - 1
. (a) 272/3 =
-- 9
10g27 9 =
2
" 3
(b) 163/" =
= 8
10gl6 8 =
I
A
12. (a) logs I = -2
3-2 = 1
(b) 49'/2 = 7
log49 7 = 5
14. y = 3^
16. y = 2^
j:
-1
0
1
2
3
y
1
9
1
3
1
3
9
18. y = 3-W
x
-2
-1
0
1
2
j:
0
±1
±2
y
16
2
1
2
16
y
1
I
3
1
9
12 3 4
20.
(a) log3
81 ^
= X
y--
1
" 81
X -
= -4
(b) logs
36 =
= X
6^ =
-- 36
X -
= 2
22. (a) log, 27 = 3
If' = 11
b = 3
(b) log, 125 = 3
fo3 = 125
b = 5
Section 5.5 Bases Other than e and Applications 517
24. (a) log3JC + log3(A:-2) = 1
logjW^c - 2)] = 1
x(x - 2) = 3'
;c2 - 2;t - 3 = 0
U + l)(x - 3) = 0
x= -10Rx = 3
j: = 3 is the only solution since the domain of the
logarithmic function is the set of all positive real
numbers.
(b) logjoU + 3) - logioj: = 1
logi
x + 3
' X
x + 3
= 10'
a: + 3 = 10a:
3 = 9jt
1
^ = 3
26. 5^ = 8320
6;cln 5 = In 8320
^ hi 8320
^~ 6 In 5
- 0.935
28. 3(5^-') = 86
5.-1 = 86
3
ix - l)ln 5 = ln(y)
^86
Ini
X- I =
In 5
;c= 1 +
In 5
= 3.085
30.
(-IT
365r ln( 1 + |^ j = In 2
t =
1
ta2
365
'"(-I)
6.932
32. log,o(f - 3) = 2.6
f - 3 = 1026
r = 3 + 102« = 401.107
34. logsVx - 4 = 3.2
4 = 53-2
;C - 4 = (53.2)2 = 56.4 ^
jc = 4 + 5*" = 29,748.593
36. /(f) = 300(1.0075'^) - 735.41
Zero: t = 10
(10.0)
38. g(x)= 1 -21og,o[xU-3)]
Zeros: x = -0.826,3.826
518 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
40. fix) = 3^
g(x) = l0g3 X
X
-2
-1
0
1
2
fix)
I
9
1
3
1
3
9
X
I
9
1
3
1
3
9
six)
-2
-1
0
1
2
42. gix) = 2--
g'ix) = -i\n2)2-
44. y = jcCe-^')
^ = 4-2(ln6)6-2') + 6-2^
dx
= 6-^-2;c(ln6) + l]
= 6-2^(1 - 2;c In 6)
32/
46. fit) = y
/'(/)
t(2 In 3) 3^ - 3^
3^(2fln3 - 1)
48. gia) = S""/- sin 2a
g'(a) = 5-"/2 2 cos 2a - lOn 5) S-"/^ sin 2a
50. y = logio (2j:) = logm 2 + logmX
dx ;clnlO a: In 10
52. /iW = log3
tV^c - 1
= logj x + - log3 (jt - 1) - logj 2
h'ix)=^^ + -
jc In 3 2 (jc - 1) In 3
-0
1
In 3
1
In 3
1_
X ^ 2{;t
3x- 2
2xix - 1)
]
54. y = logi
x^- 1
= logloU-- 1) - lOgioX
1
(iy 2x
dx ~ ix^ - l)lnlO X In 10
= 1 r 2x _ n
~ In loL^c- - 1 x\
1 r ;c^+ 1 ]
In loUU^ - 1)J
56. /(f) = r3/2 log, jm = I
3/2
1 ln(f + 1)
2 In 2
fit) =
1
21n2
[^rl
- + |/'/Mn(f + 1) I
58. )' = x'-'
Iny = (jc - l)(lnx)
rJ:-2 C„ _
ix - 1 + X In x)
60.
y = (1 + xf'^
Iny = -In(l + ;c)
1 I 1
KS=KTi7)-»»-f7
dx X
zM
xL^c +
(1 + x)^/^
InU + 1)1
"_J InU + 1)"
x+ \ X
^■\
62. 5-^^
In 5
+ C
64.
5^) A = I (27 - 25) dx
Section 5.5 Bases Other than e and Applications 519
66.
J(3-.)7'-^>^^=4j-
2(3 - x)!^^'"^' dx
2In7
£7(3 -.P] + c
S. 2
68. I 2 cos jc dx, « = sin jc, du = cos a; olx
1
In 2
2suijr + (;;
70. (a) y
,'-SWs-—- v////v\\v.
(b) "T = e^^-^cosx (-77,2)
ox
=/'
(it, 2): 2 = e^""' + C = 1 + C ^ C = 1
72. \og^x
\nx _ logio^t
In b logio b
74. /W = logioJT
(a) Domain: x > 0
(b) y = logioJ:
10>' = a:
/-•W =10'
(c) log,o 1000 = log.o 1(P = 3
Iog,olO,000 = log,o 10* = 4
If 1000 < .V < 10,000, then 3 < f{x) < 4.
(d) If/W < 0, thenO < .t < 1.
(e)/U) + 1 = log,oJt + log,ol0
= log,o(10;c)
X must have been increased by a factor of 10.
(f) 10glo(^-j = log,o-^| - log,o-t2
= in — n = 2n
Thus, jtiAj = 10^ = 100^.
76. f(x) = a^
(a) fiu + v) = a" + '' = a" a'' =f{u)f(v)
(b) f{Zx) = a^= {a^y = [f{x)Y
78. V{t) = 20,0001
(a)
3V
V{2) = 20,0001 |r = $11,250
(b)f=20,00o(ln|)(f
Whenf = 1: ^« -4315.23
When f = 4: ^ = - 1820.49
(c)
Horizontal asymptote: v ' = 0
As the car ages, it is worth less
each year and depreciates less
each year, but the value of the
car will never reach $0.
520 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
80. P = $2500, r = 6% = 0.06, t = 20
/ 0.06 N^O"
A = 2500(l +^^^j
A = 2500e(°'^**^* = 8300.29
n
1
2
4
12
365
Continuous
A
8017.84
8155.09
8226.66
8275.51
8299.47
8300.29
82. P = $5000, r = 7% = 0.07, t = 25
0.07^25"
A = 50001 1 + 1
A = 5000eO'"<25)
84. 100,000 = Pe°
P = 100,OOOe-oo«'
n
1
2
4
12
365
Continuous
A
27,137.16
27,924.63
28,340.78
28,627.09
28,768.19
28,773.01
f
1
10
20
30
40
50
P
94,176.45
54,881.16
30,119.42
16,529.89
9071.80
4978.71
0 07\3i
86. 100,000 = P\\+ ^1
P = ,00,0001 1 + H)-
t
1
10
20
30
40
50
p
93,240.01
49,661.86
24,663.01
12,248.11
6082.64
3020.75
88. LetP = $100, 0 < r < 20.
400
(a) A = lOOfiOo^r
A(20) = 182.21
(b) A = lOOe""^'
A(20)« 271.83
(c) A = lOOgOo*'
A(20) « 332.01
92. (a) 12.000
(b) Limiting size: 10,000 fish
, , 10,000
(c) p(t) =
p\t) =
~ (\ + 19e-'/5)2
p'{\) = 113.5 fish/month
p'(10) » 403.2 fisii/month
1 + 19e-'/5
e-'/i /19
(1 + 19e'/5)2l 5
38,000e-"'5
: 10,000)
90. (a) lim
0.86
= 0.86 or
(b)
-0.86(-0.25)(e-°-^") 0.2156-"-^"
(1 + e-o-zs")
(1 + e-o-25")
P'(3) - 0.069
F'(IO) - 0.016
,. , 38,000, ,,,
(d) p'tr) = -r-{e-''^)
1 - 19e-'/^
(1 + 19e-'/=)3
h
19e-'/5 = 1
t
In 19
f = 5 In 19 = 14.72
Section 5.5 Bases Other than e and Applications 521
94. (a) ^'i = 6.0536;c + 97.5571
y^ = 100.0751 + 17.8148 In X
Vj = 99.4557(1.0506)^
y^ = 101.2875xO"'-'i
(b) 150
100
>3 seems best.
(c) The slope of 6.0536 is the annual rate of change in
the amount given to philanthropy.
(d) For 1996, .« = 6 and y,' = 6.0536, >',' = 2.9691,
j3' = 6.6015,^4' -3.2321.
^3 is increasing at the greatest rate in 1 996.
96. A = S-'dx = [-^1 = -^ - 23.666
Jo Lin 3 Jo In 3
98.
X
1
10-'
10-2
10-^
io-«
(1 + x)'"^
2
2.594
2.705
2.718
2.718
100.
f
0
1
2
3
4
y
600
630
661.50
694.58
729.30
y = ciJd)
When f = 0, >- = 600
y = 600{kf)
C = 600.
630^^05 66yO 69i58»,io5 ^^^ «= 1 05
600 ' 630 ' 661.50 ' 694.58
Let/t = 1.05.
y = 600(1.05)'
102. True.
/(e" + ') -f{e") = Ine"*' - In e"
= n + \ - n
= 1
104. True.
cPy _
dx" '
Ce'
yforn = 1,2,3,
106. True.
f(x) = g(x)e' = 0 ^
g(x) = 0 since e' > 0 for all .r.
522 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
108. y = x^^''
y' l^\
— = sinx — + cos;c • Injc
y ' = -t^"-' — - + cos X • In X
Tangent line: y — — = IIjc — —
y^x
Section 5.6 Differential Equations: Growth and Decay
1 '^y A A dy , Jx
2
dy _
dx
= 4 -
- y
dy
A-y
-dx
■
'^ dv
= -
-dx
, 4
-/>
ln|4-
- y\dy -
= —x
+ c.
A - y -
= «-•>
■+c, =
Ce
y =
= 4-
- Ce"^
/3yy^ = |,
/x dx
2 3 '
8. y'-
= j:(i + y)
y'
1 +.V
= ;c
f/' dx-
= X dx
Ji+y
= xdx
ln(l + y) =
=f-.
l+y-
= g(rV2) + C,
y --
= gC, ^/2 _ 1
= C^/2 - 1
9y2 _ 4^/2 = c
10. xy + y' = 100;c
y' = 100;c + xy = ^100 - y)
y' ^
100 - > "^
jTo^'^^^r'^"
-ln(100-y)=y + C,
ln(100 - y) = -| - C,
100 - >- = e-0^/2)-c,
-y = e-c. e-^/2 - 100
>- = 100 - Ce-^/^
Section 5.6 Differential Equations: Growth and Decay 523
12. ^ = /t(10 - t)
dt
dp ^
—-dt =
dt
-
= k(\Q -
- t)dt
dP--
->■
-ty- + c
P =
^>-
-tY + c
14.
dx'
= kx{L - y)
1
L -
dy
ydx
= kx
1
L-y
dx
= fccdx
1
, L -
— rfv =
V
2 ^^'
-ln(L
-y)-
L - y = e-(*^/2)-c,
-y = -L + g-C, g-tar'/2
y = L - Ce-'=^/2
16. (a)
"^^iiii
1 ! ; ' I s J '■ \ I
:oi
^ = xdx
inbl = y + c
y = e^/2+c = c,^/2
1 v.
18.
dy
dt
hi-
Vt, (0, 10) 15
■Jtdt
y = -^^1^ + C
10 = -|(0)3/2 + C ^ C = 10
^ = -2^'''' "^ '°
rfv 3
20. ^- = 4v, (0, 10)
dt 4'
In V = 7 r + Ci
4
10 = Ce° => C = 10
V = lOe^'/*
(o, 10)
22. f = ^
N = Ce'^ (Theorem 5.16)
(0, 250): C = 250
400 S
(1,400): 400 = 250e* =^ k = In-^ = In ^
When t = A,N= 250e'"=(8/5' = 250e'°<«/5>'
^25o(8V 8192
24.^ = ^
P = Ce*' (Theorem 5.16)
(0, 5000): C = 5000
(1.4750): 4750 = 5000£>* => A: = Inf^j
When f = 5. P = 5000e'"*'9/=o>(5>
= 500o(^j' - 3868.905
524 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
26. y = Ce^, (0, 4), [s, |
C = 4
y = 4e>'
, = MiM._o.4159
y = 4g -0.41591
28.
C- (3,1), (4.
5)
^ = Ce-
5 = Ce"*
2Ce« = |ce'"--
10^3* = £«
10 = e*
fc = In 10 = 2.3026
y = (7^2.3026;
5 = (7^2.3026(4)
C - 0.0005
y = O.OOOSe^^o^*'
30. y'
dy
dt
ky
-r- > 0 when v > 0. Quadrants I and II.
dx
34. Since y = Cef'^<'/2)/i62o];_ .^^,e have 1.5 = Ce['"<'''2)/i62o](iooo) => c == 2.30 which implies that the initial quantity is 2.30 grams.
When t = 10,000, we have^' = 2.30e['"'<'/2)/i62o](io,ooo) ^ o.03 gram.
36. Since y = Ce^\^(Ui)/5no\<^ we have 2.0 = Ce['"(i/2)/573o](io,ooo) =^ c = 6.70 which implies that the initial quantity is 6.70
grams. When t = 1000, we have y = 6.70et'°<i/2)/5730]{iooo) ^ 5 94 ^^^^
38. Since y = Ce^^(U2)/5iio\t^ ^e have 3.2 = Cef'-C/^'/^sojiooo =^ c = 3.61.
Initial quantity: 3.61 grams.
When t = 10,000, we have >- = 1.08 grams.
40. Since y = Ce['"<'/2)/24.360],^ ^e have 0.4 = Ce^^imvi^imm.om) =^ c = 0.53 which implies that the initial quantity is 0.53
gram. When t = 1000, we have y = o.53ef'°<'/^'/^-3«'l<"»o) = 0.52 gram.
dy fc, w
42. Since — - ky, y - Ce*^ or >- = yQe^.
po = ^0^""*
k =
In 2
5730
0.15>'o = Joe'-"" 2/5730),
(In 2)r
In 0.15 =
t = —
5730
5730 In 0.15
In 2
15,682.8 years.
44. Since A = 20,0006°"^^', the time to double is given by
40,000 = 20,000^0 055r ^^ ^e have
2 = gO.055,
In 2 = 0.055r
In 2
t =
0.055
12.6 years.
Amount after 10 years:
A = 20,000e«'055'"0' = $34,665.06
Section 5.6 Differential Equations: Growth and Decay 525
46. Since A = lO.OOOe" andA = 20,000 when ? = 5, we
have the following.
20,000 = lO.OOOeS-^
In 2
Amount after 10 years: A = lO.OOOet""^'/']"'" = $40,000
0.1386 = 13.8
48. Since A = 2000e" andA = 5436.56 when t= 10. we
have the following.
5436.56 = 2000e'O'
ln(5436.56/2000)
'■ = To
The time to double is given by
4000 = 2000e"""
In 2 ^ „,
'=aTo'^^-^^y"^'-
= 0.10 = 10%
50. 500,000 = P\l +
0.06Y'a(-'o)
12
P = 500,000(1.005)-''8o « $45,631.04
52. 500,000 = Pll +
P = 500,000 1 +
== $53,143.92
0.09 V '^'<^'
12
0.09
12
54. (a) 2000 = 1000(1 + 0.6)'
2 = 1.06'
ln2 = rlnl.06
In 2
t =
In 1.06
= 11.90 years
(b) 2000 = 1000( 1 +
2 = I 1 +
0.06
12
0.06^ '2'
12
ln2 = 12rln 1 +
0.06
12
t =
1
In 2
^^nfl.^f)
1 1 .58 years
(c) 2000 = lOOOl
i'^m
^ /, 0.06\365'
In 2 = 365r In 1 +
0.06
365
In 2
365
1 (^ ^ 0.06
= 11.55 years
(d) 2000 = lOOOgOo*'
2 = g0.06l
In 2 = 0.06r
In 2
' = 006°="-^^^'^^
56. (a) 2000 = 1000(1 + 0.055)
2 = 1.055'
ln2 = rlnl.055
In 2
In 1.055
= 12.95 years
(b) 2000 = 1000( 1 + ^^j
2=1 +
ln2 = 12fln 1 +
0.055 y 2'
12 J
0.055
12
t =
In 2
'H'^'-W)
= 12.63 years
/ 0 055\2''''
(c) 2000 = 1000(^1 + ^j
/, 0.055
.055 \3«'
In 2 = 365r Inl 1 +
0.055
365
t =
In 2
365
, /, 0.055
= 12.60 years
(d) 2000 = lOOOeOO"'
In 2 = 0.055r
'^-Me^'--^^^'"^
526 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
58. P= Ce'^ = Cgoo^i'
P(-\) = 11.6 = CgOo^K-i) => c = 11.9652
P= 11.9652e003"
P(10)== 16.31 or 16,310,000 people in 2010
60. P = Ce'^ = Ce-°-^'
P{-\) = 3.6 = Ce-oo*"-" => C = 3.5856
P = S.SSSee-ooo^'
P(10) == 3.45 or 3,450,000 people in 2010
62. (a) Af= 100.1596(1.2455)'
(b) N = 400 when t = 6.3 hours (graphing utility)
Analytically,
400 = 100.1596(1.2455)'
400
'■''''' = 100.1596
fin 1.2455 = In 3.9936
hi 3.9936
t =
In 1.2455
= 3.9936
~ 6.3 hours.
66. (a) 20 = 30(1 - e"*)
30e3o* = 10
ln(l/3) -ln3_ ^^,..
A?= 30(1 - e-00366r)
64. >> = Ce'', (0, 742,000), (2, 632,000)
C = 742,000
632,000 = 742,000e2*
. = iB(63|m2)_oo,o2
y - 742,OOOe-o"802'
Whenf = 4, >> = $538,372.
(b) 25 = 30(1 - e-o-0366,)
-ln6 ,„^
^ = ^00366 ^'^'^^y^
(a) 4 = 25(1 - e«") ==> 1 -
(b) 25,000 units (lim S = 25)
25
2i
25
(c) When r = 5, 5 = 14.545 which is 14,545 units.
k = Inl
-0.1744
(d) =
70. (a) R = 979.3993(1.0694)' = 979.3993eOo«^"
/ = -0.1385r* + 2.1770f3 - 9.9755f2 + 23.8513f + 266.4923
(b) 2<X»
Rate of growth = R'(t) = 65.7eOo«^"
(C) 500
(d) Pit) = ^
Section 5.7 Differential Equations: Separation of Variables 527
72. 93 = lOlog.o^^ = lOdog.o/ + 16) 74. Since ^ = % - 80)
6.7 = log.o/ => /= 10
-6.7
kdt
I A< VI I ' I
10~i6 -v'-^io- • "/ ln(y-80) = fa + C.
80= 101og,o77^ni=10(log,o/+ 16)
/7^* = l
= log,o 1^1= IQ-* When f = 0, >> = 1500. Thus, C = In 1420.
;nf = \,y = 1120. Thus,
k(\) + In 1420 = ln(1120- 80)
Percentage decrease: ( '»';;:J°")(100) ==95% When r = 1, , = 1 120. Thus
104
k = In 1040- In 1420 = In 7^.
142
Thus, .V = i420e['"<'*'/"»2)]' + 80.
When t = 5,y-= 379.2°.
76. True 78. True
Section 5.7 Differential Equations: Separation of Variables
Ixy
2. Differential equation: y = ^-^ — ^
X y
Solution: x^ + y^ = Cy
Check: 2x + lyy' = Cy' • '
-2x
y
(2y - C)
, _ -~2xy _ —Ixy _ —2xy Ixy
^ " 2>'2 - Cy ~ 2y2 - U^ + y^) " y^ - ^2 - ^2 J ^2
4. Differential equation: y" + 2y ' + 2y = 0
Solution: y = C[e~'cosx + C-,e~^sinjc
Check: y' = -(Cj + Cj)e-^sinA: + (-C, + C2)e"^cosx
y" = 2 Cif"^ sin x — IC-^e'^ cos .x
y"+ 2y' + 2y = 2C,<'--"^ sin a- - 2C2e--' cos .« +
2(-(Ci + C.)e-^^mx+ (-Ci + C^-'^cosx) + 2(C,e-"^cos.x + Qe""^ sin .r)
= (2C, - 2Ci - 2C2 + 2C2)e--'sin.»: + (-2C2 - 2Ci + 2C2 + 2Ci)f "'^cos.v = 0
6. y = |(e^2A- _,_ ^)
y' = f(-2e-^ + e")
y"=f(4e-^ + e')
Substituting, y" + 2y' = f(4e--t + ^t) + 2(5)(-2e"^ + e') = 2e■^
528 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
In Exercises 8-12, the differential equation is ^'''* — 16y = 0.
8. y = 3 COS Zx
;y(4) = 48 COS 2x
■y(4) _ i6y = 48 cos It - 48 cos 2x = 0, Yes.
10.
«(4)
>> = 5 In X
30
v('') = -:
30
I6y = — T- 801n;c^ 0, No.
12. y = 3e^ — 4 sin 2jc
yW = 48e^ - Msinlx
yW - i6>. = (24e^ - 64 sin 2x) - 16(3e2» - 4 sin 2r) = 0, Yes
In 14-18, the differentia] equation isxy' — 2y = x^e^.
14. V = ;cV, y' = xV + Ixe' = e^(;c2 + 2x)
xy' - 2y = x{e^{x^ + 2x)) - 2{x^e^) = x^e^, Yes.
16. >> = sin X, y ' = cos j:
xy ' — 2^ = x(cos x) — 2{sin jc) ¥" x'e^. No.
18. y = xV - ix-,y' = ;cV + 2xe^ - 10a:
xy' -ly = ;c[a:V + Ixe - lOx] - 2[.tV - S.t^] = jt^e^, Yes.
20. v = A siB (Jit
d^ , , .
-TT = —Aar sin oi/
Since (d^/dt^) + 16>' = O, we have
—Aui^ sin cot + 16A sin a)t = 0.
Thus, or = 16 and oj = ±4.
22. 2x^ - y^ = C passes through (3, 4)
2(9) -16 = C=^C = 2
Particular solution: 2x' — y^ = 2
24. Differential equation: yy' + x = 0
General solution: x^ + y~ = C
Particular solutions; C = 0, Point
C= 1,C = 4, Circles
26. Differential equation: 3a: + 2yy ' = 0
General solution: 3x^ + 2y^ = C
6x + 4yy' = 0
2(3x + 2yy') = 0
3x + 2yy' = 0
Initial condition:
yd) -= 3: 3fl)2 + 2(3)^ = 3 + 18 = 21 = C
Particular solution: 3x^ + 2y- = 21
Section 5.7 Differential Equations: Separation of Variables 529
28. Differential equation: xy" + y' = 0
General solution: >> = C, + Cj In j:
y'=C,ny"=-Cn
xy"+y' = x[-C,^] + C^ = 0
Initial conditions: y(2) = 0, >> '(2) = -
0 = C, + C2 In 2
2 2
C, = 1, C, = -In 2
30. Differential equation: 9y" - 12y' + 4>' = 0
General solution: y = e-^/^(Ci + C,j:)
Particular solution: y = — In 2 + In jc = In -
y"= l^^/^ff C, + Q + Iq.t) + ^^4C2 = \eM\c, + 2C, + \c.x
3 V3
3 V3
<^"
9y"- 12y' + 4y = 9(^e^/3 j/|q + IC. + 1^.^] - 12(e^''3)(|Ci + Q + jQa:) + 4{e^/3)(C| + Qj) = 0
Initial conditions: y(0) = 4, y(3) = 0
0 = e\C^ + 3C2)
4 = (1)(C, + 0) ^ C, = 4
0 = e-(4 + 3C2) => C2 = -J
Particular solution: y = e^l\ 4 — — x
32. ^ = .x3 - 4;c
dx
V
y = I (.x3 - 4.t) dr = — - 2x2 + C
34.
dx: ~ 1 + e'
Jt^
e"
dx = ln(l + e^) + C
1/: "^
36. -— = xcosx-
dx
y = \x cos(.r-) dx = — sin(.r-) + C
(u = x~, du = 2xdx)
38. -7- = tan- x = sec- .r — 1
dx
I (sec- X —
1) (ix = tan .r — .r -1- C
dv
40. -j- = xV5 - X. Let « = V5 - x, «= = 5 - x. dx = - 2u dw
dx
y = X
= xVS^^dr = (5 - u-)u(-2u
(-10«= + 2«'')du
3 5
-j(5 - x)V2 + |{5 - x)^/- + C
530 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
42.
dx
y = '■
5e-V2Qrj = 5(-
'2)je-
-i
= - lOe-^/2
+ c
46,
dr
' ds
dr =
r =
0.055
0.055 ds
0.0255^ + C
48. xy'^y
If-f
dx
X
44.
dy _x^ + 2
dx 2>y^
|3y2rfy = J
{x^ + 2)dx
f = ^ + 2x + C
\ny = \a X + \n C = \n Cx
y = Cx
dy
50. y-r- = 6 cos vx
dx
h^\
6 cos TTX dx
/ 6 .
-r = — sin TTX + C,
2 77 '
>^ = — smTix + C
TT
,dy
52. V^?^^^ = 5jc
dx
/-/
5;c
■ dx
Jx^- 9
54. 4y^ = 3e^
Wydy = pe^dx
2y2 = 3e^ + C
56. Jx+ Jyy' = 0
Ll/2^y = - Ll/2^
|y3/2 = _|^/2 + c,
y/2 + ^/2 = c
Initial condition: ^(1) = 4,
(4)3/2 + (1)3/2 = 8 + 1 = 9 = C
Particular solution: y'l'^ + x'l'^ = 9
58. Ixy' - \nx^ = 0
l<
60. yVl -a:^T- = -^v1 -y^
ox
(1 _y2)-l/2y^y = \{l-X^)-'^^xdx
-(1 -/)l/2= _(1 -;c2)l/2+ c
y(0) = 1: 0=-l + C=>C=l
yr^= yn^
1
dx
= 21nx
dy
■ln;c
X
^
y '■
(lnx)2
2
+ c
yd)
= 1:1--
= C
y =
)^\^xY-
4- 2
62.
ds
-2s
H^l
= e-2^rf5
-g-'' = --e"^ + C
^(0) = 0: -1 = -| + C^C=-|
1,1
2 2
g-r = i -2r + i
2 2
-r=ln|.- + i| = lnl
r = lnl
1 + e-^\
2 ;
1 +e-
Section 5.7 Differential Equations: Separation of Variables 531
64. dT + k(T - 70) dt = 0
dT
ln(r- 70) = -kt + Ci
r - 70 = Ce-*'
Initial condition: T(0) = 140;
140 - 70 = 70 = Ce" = C
Particular solution: T- 70 = 70e-*', T= 70(1 + e~*')
66.
ir 3jt
In y = in x^ + In C
Initial condition: y{S) = 2, 2' = CCS^), C =
Particular solution: Sv^ = x^, y
±rV3
68. m - — - -
dx X — 0 X
n-\
dx
X
In >> = In JT + C, = Tn jr + In C = In Cc
y = Cx
70. /U, y) = .r3 + BxV - 2y2
/(rx, ty) = t^x^ + Ifx^ - Ifr
Not homogeneous
72. /U,y) =
xy
r;if fy
r-xy
xy
tjx- + y2 7.^2 + y2
Homogeneous of degree 1
74. /U, y) = tanU + y)
f(tx, fy) = tan(fx + ty) = tan[f(.x + y)]
Not homogeneous
76. /(.r,y) = tan^ ■
ty V
f(tx, rv) = tan — = tan -
tx X
Homogeneous of degree 0
78. y
.ty
xy2 dy = {:c' -\- y") dx
y = vx, dy = X d\) -\- V dx
x{vxY{x dv + V dx) = {x^ + (vxY) dx
X* v^ dv + x^ v^ dx = x^ dx + v^ x^ dx
xv^ dv = dx
"l
Jv2.v = /i
dx
- = ln|x| + C
(^y = 3 ln|.t| + C
80.
X- + y-
2xy
, y = vx
dv X' + v-x-
V + X — =
dx 2x- V
2v dx + 2x dv = dx
1:3^- -/f
ln(v2- 1) = -Inx + lnC= In-
v2- 1 =
4-1 = ^
X- X
y- — -ir = Cx
532 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
, 2y + 3y
82. y' = -,y = vx
dv 2x + 3vx
V + x-r = = 2 + 3v
dx x
dx
/if. -If
ln|l + v| = \nx^ + \nC = \nx'^C
l+v=x^C
X
X
y = Cy? — X
84. — y^ etc + j:U + y) rfy = 0, >> = vx
-x'^v^dx + {x'^ + x^v){vdx + xdv) = 0
/^— If
V + In V = - In ;c + In Ci = In
Ci
1 ^1
V = In —
XV
VX
y
y = Ce-y^''
Initial condition: ^(l) =1,1 = Ce~' =^ C = e
Particular solution: y = e'~^/^
86. (2jc2 + y2) dx + xydy = 0
Let y = v;c, ofy = j: dv + V tic.
(2x' + vV) dx + x(vx){x dv + V dx) = Q
(2^2 + 2^2 V 2) dx + x^v dv = Q
(2 + lv^)dx = -j:vrfv
;c 1 + v'
1
rfv
-21nx = -ln(l + v2) + C;
ln;c-2 = ln(l + v^y^ + In C
x-2 = C(l + v2)'/2
- = C(x2 + /)l/2
X
y{\) = 0: 1 = C(l + 0) => C = 1
-= y?T7
88- T
ax
+ Q
3/2 + ^2
1 = xjx^ + f
Section 5.7 Dijferential Equations: Separation of Variables 533
90. ^ = 0.25x(4 - y)
ax
dy
A-y
025x dx
\^>-h
25xdx
1
\jxdx
1
In b - 4| = --x^ + C,
;y = 4 + Ce-^um
92. £ = 2-j.,>.(0) = 4
I I I I
\ \ \ \V\
\ \ \ \
\ \ \ \
(111
\ \ \ \ \
\ \ \ \ \
\ \ \ \ \
N S N V
/./,/./ /././././.
I I I I l\l I I I I
94. ^ = 0.2jc(2 - y), y(0) = 9
96. ^ = Ay, :y = Ce*'
dt
Initial conditions: ^(O) = 20, y(l) = 16
20 ^ Ce° = C
16 = 20e*
A: = ln|
Particular solution; y = 20e"°<''/5'
When 75% has been changed:
5 = 20e"°(^/5'
1 = grln(4/5)
ln(l/4)
ln(4/5)
98. ^ = /t(x - 4)
dx
The direction field satisfies {dy/dx) = 0 along x = 4:
Matches (b).
6.2 hr
100. ^ = ky""
dx
The direction field satisfies (dy/dx) = 0 along y = 0, and
grows more positive as y increases. Matches (d).
102. From Exercise 101,
w = 1200 - Ce-^ k = 1
w = 1200 - Ce-'
w{0) = H'o = 1200 - C ^ C= 1200 - Ho
w = 1200 - (1200 - Wo)e-'
534 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
104. Let the radio receiver be located at {xq, 0).
The tangent line toy = x - j:^ joins (- 1, 1)
and (xq, 0).
(a) If [x, y) is the point of tangency on the >> = x - x^,
then
1 - 2;c =
y— l_x-x^— 1
X + I X + 1
X — 2x^ +1— 2jc = .r — j:-— 1
a:2 + It - 2 = 0
-2 ± 74 + 8
= -1 + 73
Then
y = x-x"- = 373 - 5
1-0 1 - 373 + 5 6 - 373
l--to -1 + 1-73 -73
73 = (1 + Xo){6 - 373)
= 6 - 373 + xg(6 - 373)
473 - 6
° 6-373
= 1.155
(c) 10
(b) Now let the transmitter be located at (- 1, h).
1 o y- h _ X- x^ - h
'-^ = rri- x+1
x-2x^+\-2x = x-x^-h
x^ + 2x-h-l=0
(-2 + 74 + Mh + 1))
1 + 72 + ;z
y = j: — x^
Then,
= 372 + h- h- 4
0 /; - (372 + h~ h- A)
■1--X0 -1 - (-1 + 72 + /;)
2/! + 4 - 372 + /i
-72 + /i
Xo+ 1
72 + /!
2/! + 4 - 372 + h
hjl + h
2h+ A- 372 + h
- 1
There is a vertical asymptote at /i =5, which is the
height of the mountain.
106. Given family (hyperbolas): x^ — 2)r = C
Ix - Ayy' = 0
X
y =
2y
Orthogonal trajectory: y' =
-ly
n-\i
dx
\ny = -21nA: + \nk
y = foc-2 =
108. Given family (parabolas): y^ = 2Cx
2yy'=2C
' = £ = yl(l\=y.
^ y 2x\yj 2x
Orthogonal trajectory (ellipse):
2x
Jyrfy=-J:
2xdx
= -x^ + Ki
2x'^ + y^ = K
J^
^
Section 5.8 Inverse Trigonometric Functions: Differentiation 535
110. Given family (exponential functions): y = Ce^
y' = Ce' = y
Orthogonal trajectory (parabolas): y' = —
\ydy= -\dx
y- = -2x + K
112. The number of initial conditions matches the number of
constants in the general solution.
114. TWo families of curves are mutually orthogonal if each
curve in the first family intersects each curve in the
second family at right angles.
116. Tnie
dy
^=U-2)Cv+l)
118. True
X' + y^ = 2Cy
dy X
dx C - y
X K - X Kx
-x^
x2 + y2 = 2Kx
dy ^K- X
dx V
2Kx - Ix- _ X- +y^- 2x^ _ y^
C — y y Cy — y^ 2Cy — ly'^ x''- + y — 2y x- — _v'
= -1
Section 5.8 Inverse Trigonometric Functions: Differentiation
2. y = arccos x
(a)
x
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
y
3.142
2.499
2.214
1.982
1.772
1.571
1.369
1.159
0.927
0.634
0
(c)
(d) Intercepts: I 0, — ) and (1, 0)
No symmetry
^-f
3 '
-f)
(-^._). -v^.-f
6. arcsin 0 = 0
536 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
8. arccos 0 = —
2
10. arc cot(- J3) = -^
6
12. arccos(^-— j = -
14. arcsin(-0.39) = -0.40
16. arctan(-3) = -1.25
18. (a) tanl arccos ^r- j = tanl — 1 = 1
(b) cos arcsin
5\ 12
13/ 13
20. (a) sec arctani -7) =
5
vS
-3
(b) tan
syiT
arcsinl-- = ^^
22. y = sec(arc tan 4x)
8 = arctan 4x
■y = sec e = Vl + I6x^
24. y = cos(arccotx)
6 - arccot x
y = cos 6 =
vO^+T
26. y = sec[arcsin(;c ~ 1)]
9 = arcsin(.r — 1)
1
>> = sec 6 =
Jlx - x^
( . X- h\
28. y = cosi arcsin I
6 = arcsin
X- h
y = cos 0 =
Vr^ - U - /z)-
Vr^-U-Zl)^
30.
v^
^,
32. arctan(2x - 5) = - 1
2x - 5 = tan(-l)
1,
X = -(tan(-l) + 5) = 1.721
Asymptote: x = 0
arccos r = 0
cos 6 =
tan e =
74^-^
Section 5.8 Inverse Trigonometric Functions: Differentiation 537
. arccosx
—
arcsecj:
X
=
cos(arcsec x)
X
=
x
x^
=
1
X
^
±1
V?^\
36. (a) arcsm(— x) = — arcsin^, \x\ < 1.
Let>' = arcsin(— j:). Then,
— ^=sin3' => x = — siny => x = sin(— y).
Thus, —y = sicsmx =^ y = — arcsin j:. Therefore,
arcsin(-j:) = -arcsinx.
(b) arccos(-.r) = tt - arccos x, |j:| < 1.
Lety = arccos(-.r). Then,
—X = cosy =^ X = — cosv => ;c = cos(7r — y).
Thus, TT — y = arccos x =^ y = tt - arccos x.
Therefore, arccos(-.i:) = tt — arccos x.
38. f(x) = arctan x +
X = tanly —
Domain: (— oo, oo)
Range: (0, tt)
f(x) is the graph of arctan x shifted ir/l units upward.
42. f(t) — arcsin t-
2t
f\t)
40. fix) = arccosl —
— = cos y
4
X = A cos y
Domain: [-4,4]
Range: [0, tt]
44. f(x) = arcsec 2x
2
VT^
fix) =
\2x\ V4x- - 1 \x\j4x~- 1
c^^-_
o*
^^r /,..-]
<y:f
d
/i-t^+<«i^''7
A/' '
-V,
46. /W = arctan Vx
fix)
1 + xj\2j'xl 2v/x(l + ;c)
50. f(x) = arcsin a: + arccos x =
/'W = 0
48. h{x) = x^ arctan x
h \x) = 2x arctan x +
1 +x'
52. y = ]n{t- + 4) - - arctan -
It 1
f + A 2
'* 5
TV©
2t
1 2f- 1
ri + 4 r^ + 4 r + 4
i-C.rc..^7.
54. y = -
.rV'4 - .IT + 4 i
56. V = .V arctan Iv — r ln(l + \yr)
4
,_(4-^)-./.(-^),VT^,2-;,====
rfv
iv 1 + 4.T-
+ arctan(lr)
1/ 8.V
4V 1 + 4.V-
arctan(lv)
JiT^x
+ Ja^^- +
vT
74"
538 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
. X
58. y = 25 arcsin - - a:V25 - x^
60. y = arctan
1
j>'=5
-^=i_- 721-1^ -.-(25 -.r'/^(-2.)
25
(25 - x^) ^ x^
V25 - x- 725 - X- 725 - x^
2x2
725 - x^-
2 2(^2 + 4)
1 1
2 1 + (x/2)2 2
2 x
+ -(.2 + 4)-2(2x)
jc^ + 4 (;c2 + 4)2
2x^ + S + X
(x^ + 4)2
62. f(x) = arctan .r, a = 1
1
fXx) =
1 +.2
, , _ —2x
^^''' " (1 + .2)2
P,{x)=f{\)+f'{\){x-\) = ^ + \{x-\)
AW
P,(X) =/(l) +/'(1)(X - 1) + |/"(1)(X - 1)2 = J + \{X - 1) - ^(X - 1)2
64. /(;c) = arcsin X — 2x
1
/'(;c) =
7r
-2 = 0 when 7l - x^ =
1
x = ±
fix)
75
2 ■
X
(1 - .2)3/2
/"I'^'l > 0
73
Relative minimum: I ^r— , - 0.68
r(-f)<o
Relative maximum: ( — r-, 0.68
73
66. f(x) = arcsin x — 2 arctan x
1 2
fix)
71^
1 +.2
= 0
1 + 2.2 + ;c4 = 4(J _ ^2)
;d + 6.2 - 3 = 0
. = ±0.681
By the First Derivative Test, (-0.681, 0.447) is a relative
maximum and (0.681, —0.447) is a relative minimum.
68. arctan 0 = 0. tt is not in the range of >> = arctan ..
70. The derivatives are algebraic. See Theorem 5.18.
72. (a) cot e =
(b)^ =
= arccotl-
-3 dx
dt .2 + 9 dt
If.= 10,^- 11.001 rad/hr.
dt
74. cos d =
750
6 = arccos
V s
de^Mds^^ -1 (~'^^^\ ^
dt ds dt 71 - (750A)2V s^ I dt
750 ds
sjs^ - 7502 dt
If. = 3,— =66.667 rad/hr.
dt
A lower altitude results in a greater rate of change of 6.
Section 5.9 Inverse Trigonometric Functions: Integration 539
76. (a) Let y = arcsin u. Then
smy = u
cos y • y ' = u'
dy _ u'
dr cos y J\ - u^'
(c) Let y = arcsec «. Then '
secy = u
dy
sec >■ tan y — = «
ax
dy _ u' _ u'
dx secy tan y \u\Ju- - V
Note: The absolute value sign in the fomiula for
the derivative of arcsec u is necessary because the
inverse secant function has a positive slope at every
value in its domain.
(e) Let y = arccot u. Then
cot y = M
2 dy
— esc' y -— = M
dx
di
dx
- esc- y
1 + m2
(b) Lety = arctan u. Then
tany = tt
2 dy
sec'y— - = u
dx
dy _ u'
dx sec^y i + u}
(d) Let y = arccos u. Then
cosy = u
. dy
— sin y -— = u
dx
dy _ m' _
dx sin y
vr
(f ) Let y = arccsc m. Then
cscy = u
dv
— esc V cot V -p = u
' dx
dy-_
dx
-cscycoty |«|Vm- - 1
Note: The absolute value sign in the formula for the
derivative of arccsc u is necessary because the inverse
cosecant function has a negative slope at every value in
its domain.
78. f(x) = sin X
g{x) = arcsin(sinx)
(a) The range of y = arcsin x is - it/2 S y ^ tt/I.
(b) Maximum: v/l
Minimum: - n/l
80. False
The range of y = arcsin x
[IT tt]
82. False
arcsin^ 0 + arccos- 0 = 0 + | y ) ^1
Section 5.9 Inverse Trigonometric Functions: Integration
r 3 3 r 2 3 r'
, dx = -\ , dx = - arcsin(2jc) + C 4.
J VI - 4x2 2} VI - 4x- 2 I
V4
-,dx =
arcsm -
TT
6
r 4 4 r 3 4 r I r i t^
^- JtT^'^ = sJlTa?^'^ = 3 arc.an{3..) + C 8. J^^^J-^^t = [l^<^tan-J _ =
^"•/4 + (.v-l)=''"^ = I^^V")
+ C
TT
36
^•/FTT'^^ = /^--')'^^ = i-
1-3 - X + C
540 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
14. Let M = r^ du = 2t dt.
16. Let u — x^,du = 2x dx.
lAre"' = lJ(ipTW^''^* ^arctanf + C l^^J^,^ = \\
V^^^^^ 2j xVM^ - 22
(2j:)dc
1 ^^ ^
= — arcsec y + C
18. Let M = arccos x, du =
yn^
dx.
n/V2
Jo
arccos x
dx
ri/72_
Jo
yr^
-dx
—- arccos^ j:
1/72 3,
32
« 0.925
20. Let M = 1 + j:^, Jm = 2x die.
J-/3l+^' 2j_^l+.
-J2x)dx
= |-ln(l+;c2)
->/3
-In 2
^r
;dx
1(73)2
3 + U - 2)2 Ji (V3)2 + U - 2)2
-d:x:
1 x-2
— ;= arctan — 1=-
73 V 73
7377
18
J-V2
g 1 + Sin-;
dx = arctan(sin x)
7r/2
0
2L
4
26.
— dx. u = ^Jx, du = — -1=
x) 2vJc
J 27^(1 +
3 f 2m du _ (_du__ _
2J m(1 + tt2) - -^J 1 + j<2 -
ate, dx = ludu
= 3 arctan m + C
3 arctanTc + C
r 4x + 3 , , ^x f -2x , ,r
28. \-^=J=dx^ {-2)\-j=^=dx + ■i\-j^==dx = -471 -x2 + 3arcsinA:+ C
^"•Ju+l)2 + 4'^ = 2j(.+ l)2 + 4^-J(;c+l)2 + 4^
= - ln(;c2 + 2x + 5) - - arctan! —r— I + C
32.
34.
{^ dx r- dx ri /x + 2\12 1 (4
L.2 + 4.+ 13 = L(x + 2)2 + 9 = [3 =^''H~5-JJ_2 = 3 ^'^^ll
/.2?2x + 2^ = l^T^l'^ - ^/l+U+l)2'^ = ln|.2 + 2. + 2| - 7 arctan(x + 1) + C
36. , ^ ^r =
J V-;c2 + 4;c J
V4 - (x2 - 4j: + 4)
2
a!x
74 - (;c - 2):
A
. ■ /^ - 2\
= 2 arcsin — r — +
38. Let tt = j:2 - 2x, dw = (2a; - 2) dx.
= 7.x2 - 2x + C
40.
f ^* = f ,L=^- „sec|. - ,| . C
J U - l)7;c2 - 2x Jix- 1)7U -1)2-1
Section 5.9 Inverse Trigonometric Functions: Integration 541
42. ha u = x^ - A, du = 2x dx.
\
79 + 8p^r?
-.dx
^w
2x , 1 /;c2 _ 4\
')
44. Let u = Jx--1, u^ + 2 = X, 2u du = dx
I :~ dx = I -r ^M = I ; — du = l\du — 6] —:.
J;c+1 Jm2 + 3J„2 + 3 J J„2 +
c/m
= 2m ;= arctan ^
V3 ^
^ + C = 2Vj: - 2 - 2V3 arctan .^'^ + C
/3\2 9 9 9 / 3\^ 9
46. The term is ( ^ 1 = ^: x^ + 3a: = ;c2 + 3;c + ^ - ;^ = ( x + 1 1 - ^
48. (a) J e^ die cannot be evaluated using the basic integration rules.
(b) \xerdx = \e^ + C,u = ,? ^ fl
J 2 (c)Jp
^*' ^^^ J dx cannot be evaluated using the basic integration rules.
J yf^ dx = Ij :p^ & = i arctan(x^) + C, « = x^
(c) -re'/^dx = -ei/^+ Cm =
1
(b)
(c)
4^3 . 1
1 + X
^dx = - ln(l + x") + C, M = 1 + x"*
52. (a)
dy
(b) ^ = xyi6^, (0, -2)
dj
= xdx
arcsinl jl = — + C
(0, -2): arcsini --1 = C =^ C
. (y\ x' ir
2L
6
V . /at 7r\
4 = nT~?J
V = 4 sin
2 6
-f = 7>fe-<°"^
7 1 /■ / /
/ / / / /
^ / / yjf
//III
y / / / /
y y y y y
^^ y y y
N N -^ ^ ^
^ ^ ^ X X
542 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
56.
A = . dx = arcsin — = —
Jo 74^^ L 2jo 6
58.
arcsin xdtc ~ 0.571
Jo
1 P*^
60.FW=-| ^
+ 1
dt
(a) F(a:) represents the average value off{x) over the interval [x, x + 2]. Maximum at ;c = — 1, since the graph is
greatest on [— 1, 1].
-[=
(b) F(x) = I arctan f
1
F'{x) =
= arctan(x + 2) — arctan x
1 (1 + x^) - (;c^ + 4x + 5) _ -4(x+ 1)
1 + (x + 2)2 1 + x^ (x2 + l)(x2 + 4x + 5) (x^ + l)(x2 + 4x + 5)
= 0 whenx = — 1.
^' J V6x - x^ '
62. ^ ,dx
J Vox - X-
(a) 6x - x2 = 9 - (x2 - 6x + 9) = 9 - (x - 3)^
dx . /x — 3
1
J V6x — x^ J .
X- J 79 - (x - 3)2
(b) u = Vx, M~ = X, 2u du = dx
2
— arcsin — - — + C
/ti^^'"''"^^/
. du = 2 arcsin —p= ] + C = 2 arcsini
76^^ V76,
(i)
+ c
(c)
The antiderivatives differ by
a constant, 7r/2.
Domain: [0, 6]
64. Let/(x) = arctan X
1 +x2
,rY N 1 1 -x2 2x2
1 + x"' (1 + x^r (1 + X*-)
Since/(0) = 0 and/is increasing for x > 0, arctan x - ^ > 0 for x > 0. Thus,
X
arctan X >
1 +x2'
Let g{x) = x — arctan x
Since g{0) = 0 and g is increasing for x > 0, x — arctan x > 0 for x > 0. Thus, x > arctan x.
Therefore,
X
2 4 6 8 10
1 +X2
- < arctan X < x.
Section 5.10 Hyperbolic Functions 543
Section 5.10 Hyperbolic Functions
pO + pO
2. (a) cosh(0) = ^ = 1
4. (a) siiih-'(O) = 0
(b) tanh-'(O) = 0
(b) sech(l)
e + e
= 0.648
6. (a) csch-'(2) = Inl ^ "^„ | - 0.481
(b) coth->(3)=^ln(|j = 0.347
„ 1 + cosh 2x 1 + (e=^ + e"2x)/2 g2x + 2 + e'^ fe' + C'^Y
8. = = = — T— =cosh^^
fe' — e~^\le' + e~-^\ e" — e^
10. 2 sinh x cosh jc = 2 = = sinh 2x
12. 2 coshi — :H- 1 cosh( — -^ 1 = 2
= 2[^^±^
e-y + e-
gU-y)h + g-(j:-y)/2"|
e"" + e~'' e^ + e~y
cosh X + cosh y
14. tanh x = —
2
- + sech^jc = 1 ^ sech^j: = - => sech^: = -r—
2/ 4 2
cosh X =
1 273
/3/2
coth;c = :f^ = 2
sinh X = tanh x cosh j:
1V2V3\ v^
2/V 3
Putting these in order:
73
sinh X = —— csch x = Jl>
cosh jt = -^i; — sech x = — ;-
3 2
tanhj:
1
coth X = 2
csch ;c
73/3
73
16. y = coth(3x)
y' = -3csch2(3jc)
18. g(x) = In(coshjc)
1
gW
cosh X
(sinh.v) = tanh.v
20. y = X cosh X — sinh x
y' = X sinh x + cosh x — cosh .r = x sinh .v
22. /i(r) = r - coth r
/( '(0 = 1 + csch- 1 = coth- f
544 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
24. g{x) = sech- 3j:
g'{x) = -2 s&Mix) sech(3j:) tanh(3x)(3)
= - 6 sech^ 3x tanh 3x
26. fix) = 6"^"
fix) = (cosh;c)(e=''=^^)
28. >- = sechU + 1)
>> ' = — sechU + 1) tanh(jc + 1)
30. fix) = X sinhCr - 1) - coshU - 1)
fix) = j:cosh(x - 1) + sinhU - 1) - sinh(jr - 1) = xcosh(x - 1)
fix) = 0 for X = 0. By the First Derivative
Test,(0, -cosh(- 1)) = (0, - 1.543) is a relative minimum.
32. hix) = 2 tanh x - x
2
\
(0.88. 0.53)
(-0.88,-0.531
\
34. y = a cosh j:
y' = a sinh .t
>>"= a cosh.x
Therefore, y" — y = 0.
Relative maximum: (0.88, 0.53)
Relative minimum: (—0.88, —0.53)
36. fix) = cosh a; /(I) = cosh(O) = 1
fix) = sinhx /'(I) = sinh(O) == 0
fix) = coshx /"(I) = cosh(O) « 1
/'i(^)=/(0)+/'(0)(x-0)= 1
P2(;c) = 1+5^2
V
/
/>,
38. (a) >- = 18 + 25 cosh—, -25 < x < 25
(b) At X = ±25, y = 18 + 25 cosh(l) = 56.577.
At.t = 0,y = 18 + 25 = 43.
X
(c) y' ~ sinh— .At .c = 25, y' = sinh(l) = 1.175
40. Let M = VJ, d«
Let u = Vj:, d« = — 7= ^. 42. Let u = cosh x, du = sinh j; <
ijx
^= — dx = l\ cosh V^l — ^ j (it = 2 sinh VA + C J ' + sinh^ a: J cosh-^ j:
(ic =
1
1 + sinh-^ X J cosh-^ x cosh .x
= — sechx + C
+ C
44. Let u = 2x - I, du = 2 dx.
\ sech^ilx - l)dx = \\ sech^dx - 1)(2) dx
/sech2(2.-l)<i. = i/s
^tanh(2x- 1) + C
46. Let u = sech x, du =
sech' X tanh .r dlr =
/'
sech X tanh jr tic.
sech^ xi - sech j: tanh x) dx
■/
--sech^ JT + C
Section 5.10 Hyperbolic Functions 545
„ f ,, , fl + coshZx
5. I cosh'' xdx = \ c
ir sinhlr]
50.
r
Jo
V25~
:dx =
3 Jo
. 4
arcsin —
-jc + - sinh 2x + C
2 4
52.
— , ^ dx = 2 / (2) aLc = -W-
J xVl + 4jt2 J (2x)Vl + (2;c)2 ^ \
+ Vl + 4x-
|2x|
+ C
54. Let u = sinh x, du = cosh jt (ic.
I
cosh a: , . /sinhjc\ „
, ^= ate = arcsin — - — + C
V9 - sinh-jc V 3 '
56. y = tanh-'l-
^ 1
©
1 - (;c/2)2V2/ 4 - X
arcsinl I + C
58. y = sech^Hcos 2j:), 0 < .r < —
y =
-1
cos 2xVl - cos- 2a:
since sin 2x: > 0 for 0 < x < 17/4.
( — 2 sin 2r) =
2sin2x
cos2r|sin2x| cos 2x
= 2 sec 2x,
60. >- = (csch-'A:P
y' = 2 csch" ' j:|
- 2 csch ' .V
x\jr+7-/ \x\^l + x^
62. y = ;c tanh"' x + InVl - x^ = x tanh"' .r + x ln(l - x^) 64. See page 401, Theorem 5.22.
y = x\
1 -x^
=)
1 \ _ -X
+ tanh ' JC + r = tanh ' x
1 -x2
66. Equation of tangent line through P = {xg, yg)
y
- a sech-' ^ + Va- - Xg^ =
HX - Xg)
When x = 0,
y = a sech" ' '- Va- - jCn" + Va^ - JTn^ = a sech~ ' — .
a a
Hence, Q is the point [0, a sech~'(.*o/'')]-
68.
istance from P lo Q: d = y/x
o^ + (-Va^-.o^r =
■' J. 'f "'^
- = 4©'"
3-.r
9-X* 2}9-(x^r-
3 +.t2
= -±,„
3-.r
+ c
3+:r
+ c
546 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
70. Let tt = x^'\ du = -Jxdx.
72.
f ^ =f.
]{x+ 2)V.t2 + 4;c + 8 J
ate
U + 2)V(^ + 2)- + 4
= 4-P4?l^'-
74. f—
l)V2? + 4r+8
otc
= f L
J (x + \)j2{x +1)2 + 6
^J U + \)J(x + \Y +
(;c + 1)V(^ + IP + (V3)'
.^=_J_,„fVI+7U+i)i+i^^^
j: + 1
V6
76. Let tt = 2(a: - 1), du = 2dx.
f ^^f
J (x - 1)7-4x2 + 8x - 1 J
1" J 2(x - 1)V(^^-[2(x-1)?
(&= -
^In
73
73 + 7-4^2 + 8x - 1
2(x - 1)
+ C
78. y = 1:1 ^dt = \^ —,dx + 3 It -'r:^ -dx
^ }Ax-x? J 4x - x^ J (jc - 2)2 - 4
= In 4x - x^ +=T In
(x - 2) - 2
(;c - 2) + 2
+ C = lnl4x - x2| + - In
X -4
+ C
80. A = tanh 2x dx
2g2x _ g-Zx
g2i + g-2i
ip 1
2J0 £2^ + e-2^'
= [|ln{e2' + e-2-)|
= |ln(e^ + e-*)-|ln2
(it
(2)(e2^ - e-^) dx
= ln,/ 7 °= 1.654
84. (a) v(;) = -32r
82. A
I
^(fe
3 v4^^^'
U ln(x + 7x2 _ 4)
= 6 ln(5 + 721) - 6 ln(3 + Js)
(b) 5(r)
Jv(f)rff=|
(-32f)dr= -16f + C
siO) = - 16(0)2 + C = 400 => C = 400
i(r) = - 16r2 + 400
—CONTINUED—
Section 5. JO Hyperbolic Functions 547
84. —CONTINUED—
(c)
dv
dt
-32 + kv^
J 32 - fcv2 J '
Let « = V^ V, then du = Vfc dv.
1
1
:ln
Vfc 2732
Since v(0) = 0, C = 0.
V32 + yitv
V32 - V^v
= -f + C
In
-2V32/tf
- ^-2^32*;
732- Jkv
732 + Jkv
732 + 7fcv = 6-2^^2* '(732 - 7tv)
i'(7t + 7ite-2v^') = 732(e-2v^' - 1)
■ . ^(e-2v^>_ i) e>/?af
7t(e-2>'^'+ l) ' e-^'
732
v^
zif
-Jm.
'ilkt J. „-v^2itf
gVJ2A, + g
= -^tanh(732fcf)
7fc
86. Lety = arcsin(tanh j:). Then,
e" — e""^
sin y = tanh x = :- and
e' + e
e^ — e '
tan y = = sinh x.
Thus, y = arctan(sinh jc). Therefore,
arctan(sinh ;c) = arcsin(tanh ;c).
(d) lim
732
7^
tanh(732fcr)
J22
Jk'
The velocity is bounded by - 732/7^-
(e) Since /tanh(cf) dt = (1/c) In cosh(cf) (which can be
verified by differentiation), then
/32
s(t)
/
7fc
732 1
tanh(732ytr)<if
ln[cosh(732itr)] + C
7fc 732fc
= — iln[cosh(732/:r)] + C.
When r = 0,
5(0) = C
= 400 => 400 - (l/k) ln[cosh(732fcf)].
When A: = 0.01,
izW = 400 - 100 In(cosh7o32 t)
Si(f) = - 16t2 + 400.
s^it) = 0 when ; = 5 seconds.
Sjit) = 0 when r ~ 8.3 seconds
When air resistance is not neglected, it takes
approximately 3.3 more seconds to reach the ground.
88. y = sech~' j:
sech y = X
— (sechy)(tanhy)y' = 1
-1
y =
90. y = sinh^'x
sinh y = JC
(cosh y)y' = 1
1
(sechy)(tanhy) (sechy)7l - sech-y xVT
coshy 7sinh-y + 1 Vx- + 1
548 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
Review Exercises for Chapter 5
2. /(jc) = ]n(x - 3)
Horizontal shift 3 units to
the right
Vertical asymptote: x = 3
4. \n[{x^ + l){x - 1)] = lnU2 + 1) + \n(x - 1)
r 25xr' ~\
6. 3[ln;t - 2 InU^ + 1)] + 2 In 5 = 3In;c - 6Mx'^ + 1) + In 5^ = Inx^ - \n{x^ + 1)« + In 25 = In , ^ ^ ^.^
8. In X + ln(x - 3) = 0
\nx{x - 3) = 0
xix -2) = e°
x2 - 3.T - I = 0
3 ± Vn
10. h{x) = In^^^^ = Inx + ]n{x - 1) - ln(x - 2)
X - 2
.,,,11 1 x2-4x + 2
h{x) = - +
X X— 1 X — 2 x' — 3x^ + 2x
3 + 713 , . 3-713 „
X = only since < 0.
12. /(x) = ln[x(x2 - 2)2/3] = Inx + |ln(x2 - 2)
„, , 1 2/ 2x
f'(x) = - + '
X 3\x2 — 2/ 3x3 — fix
7x2-6
14. y = -jfa + te - fl ln(a + bx)]
dy _ U ab
dx b^\ a + bx) a + bx
16. y = 1 — :; In
ax a' X
1 h
= + -?[ln(a + bx) - Inx]
ax a^
(fa
aV x2/ a\a + bx xj
ax2 a2Lx(a + bx)\
1
ax2 ax{a + bx)
(g + bx) - bx 1
ax\a + bx) x\a + bx)
20. M = In X, <iM = - <fa
X
/¥*4/<'""©-=>'>'-^
18. M = x2 — \,du = Ixdx
S^^dx = ^(^^dx = |lnlx2 - 11 + C
J x2 - 1 2j x2 - 1 2 ' '
2..f^^ = f„„.,.(l)^.[i<,n,p]; = l
-/Xf
X dx
In
77
COS| "T "■ JC
)ir
= 0-,.|J=)4..2
Review Exercises for Chapter 5 549
26.
(a)
fix)
= 5x-
1
y
= 5x-
1
y + l
5
= X
x + 1
5
= y
f-Uy\
x + 1
(b)
(c) /-'(/W) ^f-\5x - 7) = ^^"^ l^^'' = X
/</-«) =/m-m—'
28. (a) fix) = x3 + 2
y = x^ + 2
Vy - 2 = ;c
V:c - 2 = y
(b)
A
^'r^
-fl
^
(c) /-'(/W) =r '(x^ + 2) = y(;c3 + 2) - 2 = x
/(/-'W) =/{4/^r^^) = {V:^^^f + 2=x
30. (a) fix) =x'^- 5,x>0
y = x^-5
Vy + 5 = x
Vx + 5 = y
/-'W = Vx + 5
(b)
^
/
(c) /"'(/W) =/-'(x^ - 5) = VCr^ - 5) + 5 = .r forx > 0.
/{/-'W) =/(Vx + 5) = {Jx + Sf -5=x
32. fix) = xVx- 3
/(4) = 4
/'W = VT^^ + ^xix - 3)-'/^
/'(4) =1 + 2 = 3
(/-')'(4)=^-;^ = |
34. /(.r) = lnx
f-'ix) = e^
(/-')'W = ^
(/-')'(0) = e°= 1
36. (a) fix) = e'-
y = e'"-'
In y = 1 - X
X = 1 — In y
y = 1 - Inx
/-'(x) = 1 - Inx
(b)
\
\^/
/-'
(c)/-'(/(x)) =/-'(«'-') = 1 - He'-')
= 1 - (1 - x) = X
/(/"'W) =/(l - Inx) = fi-ii-'^-'* = ftox = J.
550 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
38. y = 4e-
-5-4-3-2-1 ..12345
-3--
-4
-5
40. g{x) = Inl
gXx) = 1
1 + e'j
In e* - ln(l + e^) = x - ln(l + e-')
e^ 1
\ + e" \ + e'
42. h{z) = e-^'^
h\z) = -ze-^'^
44. y = 3e^3/r
Qp-3/r
y' = 3e-V'(3r2) = ^
46. f(e) =|e=ta2e
/'(e) =005 266^*" 29
48. cos n?- = xe^'
-2xsmx^ = xey^ + «>■
dx
di _ 2x sin jc^ + e^
dx ~ xey
50. Let u = -, du = — j- dx.
dx = -e'/-^ + C
52. Let « = e^ + e'^', J« = (2e^ - e'^^) d^r.
fe^ - e-^ ^ 1 r2g^ - 2g-
Je^ + e-^ 2J e^ + e-2
■dx
54. Let M = jc' + L -i" = Sx^ alx.
Lv+'aLr = 1 e
1^(2^+1^= gx=+l(3^2)^ = ^^ + l + c
= - \n{e^ + e-^) + C
56.j^^dx='-j^2e^dx
= - ln(e^ + 1) + C
58. (a), (c) 'oooo
(b) V = SOOOe-ofi', 0 < r < 5
V'(t) = -4800e-o«'
V'(l) = -2634.3 dollars/year
V'(4) = -435.4 dollars/year
60. Area = 2e^' dx
Jo
-2e-
Jo
62. g{x) = 6(2--^)
64. y = log4j:2
Review Exercises for Chapter 5 551
66. fix) = A'e^
fix) = 4^e^ + (In 4)4^?^ = 4V(1 + In 4)
X
70. /i(jc) = logs ~rj ^ '°g5 ^ ~ logsC^ ~ 1)
'''«=i;^
1 1
x j: — 1
1
In 5
xix - 1).
68. >> = ;c(4-^)
y' = 4"-' — X • 4~' In 4
72.
f2-iA
j — dt-
1
In 2
2-1 A + c
74. ? = 50 log,o!
18,000
,18,000 - h)
(a) Domain: 0 < h < 18,000
(b)
100- •
Vertical asymptote: h = 18,000
76. 2P = Pe'O'-
2 = e'O'
ln2 = lOr
In 2
/■ = — - 6.93%
(c) f = 50 log
18,000
Hl8,000-/!
8,000
18,000 - h
18,000 -/! = 18,0O0(10-'/5O)
h = 18,000(1 - lO-'/^o)
As ;!->l 8,000, t-^oo.
(d) r = 50 logio 18,000 - 50 log,(,(18,000 - h)
dt
50
78.
dh (In
10)( 18,000 -
-h)
d-'t
50
dh'' (In
10)( 18.000 -
-h)-
No critical numbers
As r increases, the rate
of chan^
eof tlie
altitude is
increasing.
>- =
^2J
^
'
^-(600) =
^2J
= 3.868
grams
80. (a)
^= -0.012y,5 > 50
ds
^/M-
aoTl'""^'^^'
When s = 50, v = 28 = Ce-'^onm
y = 28e''*-'''»^, 5 > 50
C = 2%e°^
(b)
Speed(s)
50
55
60
65
70
Miles per Gallon iy)
28
26.4
24.8
23.4
22.0
82.
dx
e'
-2r
1
+
e"
-2i
f
e"
Zr
rfx
1 r -2e--'^
"2J 1 +«--
otc
y = -- ln(l + e---^) + C
552 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
84. V ' — e^' sin j: = 0
dy , .
—- = e' sin X
dx
e-y dy = si
sin j: (it
86.
-e >■= -
cos x + Ci
1
(C= -Q)
cos j: + C
y = \r
1
= -ln|cosx + C
cosjc + C
-j- = '— (homogeneous differential equa
3{x + y)dx - xdy = 0
Let y = vx, dy = X dv + V dx.
3ix + vx)dx - x(xdv + vdx) = 0
{3x + 2vx)dx- x^dv = 0
(3 + 2v)dx = xdv
fi . r 1
3 + 2v
dv
\n\x\ = - ln|3 + 2v| + C, = ln(3 + 2v)>/2 + In Cj
;c = C2(3 + 2v)'/2
;c2 = C(3 + 2v) = c( 3 + 2(^j
jr' = C{3x + 2y) = 3C;c + 2Cy
_^x^ -3Cx
^ IC
88. ^ = /tv - 9.8
dt
(a)
1^7^ =f^
-ln|/tv - 9.8| = r + Ci
ln|fcv - 9.8| = fa + Cj
Jtv - 9.8 = e*'+c, = (2^gh
1
9.8 + Cjg*'
At ? = 0,
Vo = -(9.8 + C3) => C3 = fcvo - 9.1
V = -{9.8 + (/b'o - 9.8)6'=']
Note that k < Q since the object is moving downward.
(b) limvW=^
(c) i(f) = J|[9.8 + (A:vo - 9.8)e*']rff
9.8? + |(/fcvo - 9.8)e^ I + C
= ^ + -^(H - 9.8)e*' + C
5(0) = ^(fcvo - 9.8) + C ^ C = So - ^(fcvo - 9.8)
^(') = ^ + ;^K - 9.8)e*' + 5o - -^^vo - 9.8)
Q Kr 1
Review Exercises for Chapter 5 553
90. hix) = -3arcsin(2x)
92. (a) Let e = arccot 2
cot 0 = 2
tan(arccot 2) = tan 6
(b) Let d = arcsec V5
sec 6 = v/5
slarcsec Vs ) = cos 8 = — ;= .
94. y = arctanlx' - 1)
2x
2x
1 + U- - 1)2 ;c^ - 2jc2 + 2
96. y = X arctan e^
^ 2U + e-'V 1 + e'
98. y = V;c2 - 4 - 2 arcsec -, 2 < j: < 4
y =
X-- 4 V^2T74
V?^^ (|x|/2)V(a:/2)2- 1 Vx^'^ \x\J^^^^ \x\V^^^^ X
100. Let M = 5;c, du = 5 dx.
Jr^'^ = i/(v^)2V(5./^^^
1 5x
r arctan — ^ + C
5^3 V3
102. J.
1.1 ■« , ^
— r cit = — arctan — + C
16 + a:^ 4 4
104. I ^ -^ (it = 4 I -7=L= 0^ + :1^ I '
ix: + - (4 - x^)-^'-(-2x)dx = 4arcsin^+ J4 - x- + C
106. Let u = arcsin x, du
jr=i^-
dx.
\
arcsin j: , 1, . ^^ ^
, dx = -(arcsin x)- + C
108.
(n
Since the area of region A is 11 I sin.vrf>).
the shaded area is I arcsin.riv = — — 1 == 0.571.
Jo
110. y = xtanh-'2jc
y = x\
2r
, , , + tanh"' It = -^ + tanh-' 2jt
1 - 4x^/ 1 - 4x-
112. Let tt = .r3. du = ?,x^ dx.
I ;r(sech x^)- dx = ^\ (sech .■^)^3jr) dr = | tanh .t^ + C
554 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
Problem Solving for Chapter 5
2. (a)
smx dx = — \ smxdx
I
arccos xdx = 2\ — ] = tt
Jo
smxdx = Q
(b)
-I 1 1 \—*^x
It It 'in In
(sin j:
Jo
+ 2)dx = 2(2 tt) = An
(d)
1
^ 1 + (tanx)-^
is symmetric with respect to the
_J ^ nl\\ ^77
(tanx)v^ 2\2/ 4
4. y = O.S-' and y = 1.2-' intersect >> = j:.
>' = 2' does not intersect y = x.
Suppose y = x\s tangent to >> = o^ at (x, y).
c^ = X =^ a= x^l'.
y'=a'lna=l =s>j:ln x^l-
\z=^\'!\x=\=>x = e,a = e^l"
For 0 < (3 < e^^' ~ 1.445, the curve y = of intersects y = x.
6. (a) y = /(x) = arcsin x
sin >> = X
J-nl
Area A = | sin >> • rfy = — cos y
tt/6
V2 , V3 V3 - 72
]"" = _ V2 ^ V3 ^
Jir/fi 2 2
= 0.1589
^--^^ i)(fh§-»'^"«
I
72/2
(bj I arcsin jccic = Area(C) = ("TJ
7r\/7^
A - B
8 2 12
72 1 \ , 72 - 73
^'X-T2J+ 2
= 0.1346
—CONTINUED—
Problem Solving for Chapter 5 555
6. —CONTINUED—
(c) Area A
e^dy
3
3-1=2
AreaB= | ln;ca[r = 3(ln 3) - A = 3 In 3 - 2 = ln27 - 2 = 1.2958
(d) isny = X
Area A
riT/3
Jit/ 4
izny dy
= —In cos vl
7r/3
7r/4
= -In^ + ln^ = ln72 = |ln2
Area C = I arctan x dx = (|^j( VI) - - In 2 - (j
= ^(4V3-3)-iin2»0.6818
(1)
8. y = e'
y' = e^
y — b = e"{x — a)
y = e"x — ae" + b Tangent line
Ify = 0,
e"x = ae" — b
bx = ab - b (b = e")
X = a — I
c = a — I
Thus, a - c = a — (a - I) = I.
10. Let u = tan X. du = sec- x dx
r-r/A
Area
r'r/4
Jo sin-
1
x + 4 cos- X
J"ir/4 1
sec-jc
0 tan2.r-^4
Jo «= + 4
= [^arctang
= farctan(|)
(it
12. (a) f = >'(1 ->').v(0)=|
ln|y| - ln|l - y\ = t + C
y
In
\-y
V
= t+ C
e'+c= c,e'
1 -V
V = C,e' - yC^e'
C,e' 1
Hence, v =
1 1
4 \ + C.
1
1 -(- C.e' 1 + C^e"
C, = 3
1 + 3e-
— CONTINUED—
(b)
= y(l - y) = y - .-v-^
dy
dt
cT-y „ , ^ , „ .,- 1
-pr = V = y — 2\T => V = 0 lor v = —
dt' ' ... -2
-^> OifO< V < ^and-^ < Oif^ < V < 1.
dt- ■ 2 dt- 2 ■
Thus, the rate of growth is maximum at y = -. the
point of inflection.
556 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
12. —CONTINUED—
(c) y' = y(l -y),ym-2
1
As before, y
1 + Ce-
^(0) = 2 = :p^=>Q=4
Thus, y =
1
The graph is different:
14. (a) u = 985.93 - 985.93
(120,000)(0.095)
12
(-^r
(b) The larger part goes for interest. The curves intersect when t = 27.7 years.
(c) The slopes are negatives of each other. Analytically,
du dv
» = 985.93 -V =>^=-^
m'(15) = -v'(15) = -14.06.
(d) t = 12.7 years
Again, the larger part goes for interest.
3-3:^S7-^
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